Q: Find the slope of the normal to the curve $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$.

The given curve is $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$ Therefore, $$ \begin{aligned} \frac{d x}{d \theta} & =\frac{d}{d \theta}\left(a \cos ^{3} \theta\right) \\ & =-3 a \cos ^{2} \theta \sin \theta \\ \frac{d y}{d \theta} & =\frac{d}{d \theta}\left(a \sin ^{3} \theta\right) \\ & =3 a \sin ^{2} \theta(\cos \theta) \end{aligned} $$ Hence, $$ \begin{aligned} \frac{d y}{d x} & =\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta} \\ & =-\frac{\sin \theta}{\cos \theta} \\ & =-\tan \theta \end{aligned} $$ Now, the slope of the tangent to a curve at $\theta=\frac{\pi}{4}$ is given by, $$ \begin{aligned} \left.\frac{d y}{d x}\right]_{\theta=\frac{\pi}{4}} & =-\tan \theta]_{\theta=\frac{\pi}{4}} \\ & =-\tan \frac{\pi}{4} \\ & =-1 \end{aligned} $$ Hence, the slope of the normal at $\theta=\frac{\pi}{4}$ is given by, $$ \frac{-1}{\text { slope of the tangent at } \theta=\frac{\pi}{4}}=\frac{-1}{-1}=1 $$

Q: Find the slope of the normal to the curve $x=1-a \sin \theta$ and $y=b \cos ^{2} \theta$ at $\theta=\frac{\pi}{2}$.

It is given that $x=1-a \sin \theta$ and $y=b \cos ^{2} \theta$ Therefore, $$ \begin{aligned} \frac{d x}{d \theta} & =\frac{d}{d \theta}(1-a \sin \theta) \\ & =-a \cos \theta \\ \frac{d y}{d \theta} & =\frac{d}{d \theta}\left(b \cos ^{2} \theta\right) \\ & =-2 b \sin \theta \cos \theta \end{aligned} $$ Hence, $$ \begin{aligned} \frac{d y}{d x} & =\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-2 b \sin \theta \cos \theta}{-a \cos \theta} \\ & =\frac{2 b}{a} \sin \theta \end{aligned} $$ Now, the slope of the tangent at $\theta=\frac{\pi}{2}$ is given by, $$ \begin{aligned} \left.\frac{d y}{d x}\right]_{\theta=\frac{\pi}{2}} & \left.=\frac{2 b}{a} \sin \theta\right]_{\theta=\frac{\pi}{2}} \\ & =\frac{2 b}{a} \sin \frac{\pi}{2} \\ & =\frac{2 b}{a} \end{aligned} $$ Hence, the slope of normal at $\theta=\frac{\pi}{2}$ is given by, $$ \frac{-1}{\text { slope of the tangent at } \theta=\frac{\pi}{2}}=\frac{-1}{\left(\frac{2 b}{a}\right)}=-\frac{a}{2 b} $$

Question 1

The slope of the normal to the curve $y=2 x^{2}+3 \sin x$ at $x=0$ is

(A) 3

(B) $\frac{1}{3}$

(C) -3

(D) $-\frac{1}{3}$

Accepted Answer

The equation of the given curve is $y=2 x^{2}+3 \sin x$.

Slope of the tangent to the given curve at $x=0$ is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{x=0} & =4 x+3 \cos x]_{x=0} \

& =0+3 \cos 0 \

& =3

\end{aligned}

$$

Hence, the slope of the normal to the given curve at $x=0$ is

$$

\frac{-1}{	ext { slope of the tangent at }(x=0)}=\frac{-1}{3}

$$

Thus, the correct option is $\mathbf{D}$.

Question 2

Find the slope of the tangent to the curve $y=3 x^{4}-4 x$ at $x=4$.

Accepted Answer

The given curve is $y=3 x^{4}-4 x$

Then, the slope of the tangent to the given curve at $x=4$ is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{x=4} & \left.=\frac{d}{d x}\left(3 x^{4}-4 xight)ight]_{x=4} \

& \left.=12 x^{3}-4ight]_{x=4} \

& =12(4)^{3}-4 \

& =12(64)-4 \

& =764

\end{aligned}

$$

Question 3

Find the slope of the tangent to the curve $y=\frac{x-1}{x-2}, x 
eq 2$ at $x=10$.

Accepted Answer

The given curve is $y=\frac{x-1}{x-2}$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(\frac{x-1}{x-2}ight) \

& =\frac{(x-2)(1)-(x-1)(1)}{(x-2)^{2}} \

& =\frac{x-2-x+1}{(x-2)^{2}} \

& =\frac{-1}{(x-2)^{2}}

\end{aligned}

$$

Now, the slope of the tangent to the given curve at $x=10$ is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{x=10} & \left.=\frac{-1}{(x-2)^{2}}ight]_{x=10} \

& =\frac{-1}{(10-2)^{2}} \

& =\frac{-1}{64}

\end{aligned}

$$

Question 4

Find the slope of the tangent to curve $y=x^{3}-x+1$ at the point whose $x$-coordinate is 2 .

Accepted Answer

The given curve is $y=x^{3}-x+1$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{3}-x+1ight) \

& =3 x^{2}-1

\end{aligned}

$$

Now, the slope of the tangent at the point where the $x$-coordinate is 2 is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{x=2} & \left.=3 x^{2}-1ight]_{x=2} \

& =3(2)^{2}-1 \

& =12-1 \

& =11

\end{aligned}

$$

Question 5

Find the slope of the tangent to curve $y=x^{3}-3 x+2$ at the point whose $x$-coordinate is 3 .

Accepted Answer

The given curve is $y=x^{3}-3 x+2$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{3}-3 x+2ight) \

& =3 x^{2}-3

\end{aligned}

$$

Now, the slope of the tangent at the point where the $x$-coordinate is 3 is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{x=2} & \left.=3 x^{2}-3ight]_{x=3} \

& =3(3)^{2}-3 \

& =27-3 \

& =24

\end{aligned}

$$

Question 6

Find the slope of the normal to the curve $x=a \cos ^{3} 	heta, y=a \sin ^{3} 	heta$ at $	heta=\frac{\pi}{4}$.

Accepted Answer

The given curve is $x=a \cos ^{3} 	heta$ and $y=a \sin ^{3} 	heta$

Therefore,

$$

\begin{aligned}

\frac{d x}{d 	heta} & =\frac{d}{d 	heta}\left(a \cos ^{3} 	hetaight) \

& =-3 a \cos ^{2} 	heta \sin 	heta \

\frac{d y}{d 	heta} & =\frac{d}{d 	heta}\left(a \sin ^{3} 	hetaight) \

& =3 a \sin ^{2} 	heta(\cos 	heta)

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)}=\frac{3 a \sin ^{2} 	heta \cos 	heta}{-3 a \cos ^{2} 	heta \sin 	heta} \

& =-\frac{\sin 	heta}{\cos 	heta} \

& =-	an 	heta

\end{aligned}

$$

Now, the slope of the tangent to a curve at $	heta=\frac{\pi}{4}$ is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{	heta=\frac{\pi}{4}} & =-	an 	heta]_{	heta=\frac{\pi}{4}} \

& =-	an \frac{\pi}{4} \

& =-1

\end{aligned}

$$

Hence, the slope of the normal at $	heta=\frac{\pi}{4}$ is given by,

$$

\frac{-1}{	ext { slope of the tangent at } 	heta=\frac{\pi}{4}}=\frac{-1}{-1}=1

$$

Question 7

Find the slope of the normal to the curve $x=1-a \sin 	heta$ and $y=b \cos ^{2} 	heta$ at $	heta=\frac{\pi}{2}$.

Accepted Answer

It is given that $x=1-a \sin 	heta$ and $y=b \cos ^{2} 	heta$

Therefore,

$$

\begin{aligned}

\frac{d x}{d 	heta} & =\frac{d}{d 	heta}(1-a \sin 	heta) \

& =-a \cos 	heta \

\frac{d y}{d 	heta} & =\frac{d}{d 	heta}\left(b \cos ^{2} 	hetaight) \

& =-2 b \sin 	heta \cos 	heta

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{\left(\frac{d y}{d 	heta}ight)}{\left(\frac{d x}{d 	heta}ight)}=\frac{-2 b \sin 	heta \cos 	heta}{-a \cos 	heta} \

& =\frac{2 b}{a} \sin 	heta

\end{aligned}

$$

Now, the slope of the tangent at $	heta=\frac{\pi}{2}$ is given by,

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{	heta=\frac{\pi}{2}} & \left.=\frac{2 b}{a} \sin 	hetaight]_{	heta=\frac{\pi}{2}} \

& =\frac{2 b}{a} \sin \frac{\pi}{2} \

& =\frac{2 b}{a}

\end{aligned}

$$

Hence, the slope of normal at $	heta=\frac{\pi}{2}$ is given by,

$$

\frac{-1}{	ext { slope of the tangent at } 	heta=\frac{\pi}{2}}=\frac{-1}{\left(\frac{2 b}{a}ight)}=-\frac{a}{2 b}

$$

Question 8

Find the points at which tangent to the curve $y=x^{3}-3 x^{2}-9 x+7$ is parallel to the $x$-axis.

Accepted Answer

The given curve is $y=x^{3}-3 x^{2}-9 x+7$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{3}-3 x^{2}-9 x+7ight) \

& =3 x^{2}-6 x-9

\end{aligned}

$$

Since tangent is parallel to the $x$-axis if the slope of the tangent is zero.

Hence,

$$

\begin{aligned}

& 3 x^{2}-6 x-9=0 \

& \Rightarrow x^{2}-2 x-3=0 \

& \Rightarrow(x-3)(x+1)=0 \

& \Rightarrow x=3 	ext { or } x=-1

\end{aligned}

$$

When, $x=3$

Then,

$$

\begin{aligned}

y & =(3)^{3}-3(3)^{2}-9(3)+7 \

& =27-27-27+7 \

& =-20

\end{aligned}

$$

When, $x=-1$

Then,

$$

\begin{aligned}

y & =(-1)^{3}-3(-1)^{2}-9(-1)+7 \

& =-1-3+9+7 \

& =12

\end{aligned}

$$

Thus, the points at which the tangent is parallel to the $x$-axis are $(3,-20)$ and $(-1,12)$.

Question 9

Find a point on the curve $y=(x-2)^{2}$ at which the tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$.

Accepted Answer

If a tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$ Then, the slope of the tangent $=$ the slope of the chord.

Hence, the slope of chord is $\frac{4-0}{4-2}=\frac{4}{2}=2$

Now, the slope of the tangent to the given curve is,

$$

\frac{d y}{d x}=2(x-2)

$$

Since, the slope of the tangent $=$ the slope of the chord.

Hence,

$$

\begin{aligned}

& 2(x-2)=2 \

& \Rightarrow x-2=1 \

& \Rightarrow x=3

\end{aligned}

$$

When, $x=3$

Then,

$$

\begin{aligned}

y & =(3-2)^{2} \

& =1

\end{aligned}

$$

Hence, the point on the curve is $(3,1)$.

Question 10

Find the point on the curve $y=x^{3}-11 x+5$ at which the tangent is $y=x-11$.

Accepted Answer

The given curve is $y=x^{3}-11 x+5$

The equation of the tangent to the given curve is $y=x-11$

Therefore, slope of the tangent is 1 .

Now, the slope of the tangent to the given curve is,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{3}-11 x+5ight) \

& =3 x^{2}-11

\end{aligned}

$$

Hence,

$$

\begin{aligned}

& 3 x^{2}-11=1 \

& \Rightarrow 3 x^{2}=12 \

& \Rightarrow x^{2}=4 \

& \Rightarrow x= \pm 2

\end{aligned}

$$

When, $x=2$

Then,

$$

\begin{aligned}

y & =(2)^{3}-11(2)+5 \

& =8-22+5 \

& =-9

\end{aligned}

$$

When, $x=-2$

Then,

$$

\begin{aligned}

y & =(-2)^{3}-11(-2)+5 \

& =-8+22+5 \

& =19

\end{aligned}

$$

Thus, the points are $(2,-9)$ and $(-2,19)$.

Question 11

Find the equation of all lines having slope -1 that are tangents to the curve $y=\frac{1}{x-1}, x 
eq 1$.

Accepted Answer

The given curve is $y=\frac{1}{x-1}, x 
eq 1$

The slope of the tangents to the given curve is given by,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(\frac{1}{x-1}ight) \

& =\frac{-1}{(x-1)^{2}}

\end{aligned}

$$

The slope of the tangent is -1

So, we have:

$$

\begin{aligned}

& \frac{-1}{(x-1)^{2}}=-1 \

& \Rightarrow(x-1)^{2}=1 \

& \Rightarrow x-1= \pm 1 \

& \Rightarrow x=2,0

\end{aligned}

$$

When, $x=0, \Rightarrow y=-1$ and when $x=2, \Rightarrow y=1$

Thus, there are two tangents to the given curve having slope -1 and passing through the points $(0,-1)$ and $(2,1)$.

Hence, the equation of the tangent through $(0,-1)$ is given by,

$$

\begin{aligned}

& y-(-1)=-1(x-0) \

& \Rightarrow y+1=-x \

& \Rightarrow y+x+1=0

\end{aligned}

$$

The equation of the tangent through $(2,1)$ is given by,

$$

\begin{aligned}

& y-1=-1(x-2) \

& \Rightarrow y-1=x+2 \

& \Rightarrow y+x-3=0

\end{aligned}

$$

Thus, the equations of the required lines are $y+x+1=0$ and $y+x-3=0$.

Question 12

Find the equations of all lines having slope 2 which are tangent to the curve $y=\frac{1}{x-3}, x 
eq 3$.

Accepted Answer

The given curve is $y=\frac{1}{x-3}$

The slope of the tangents to the given curve is given by,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(\frac{1}{x-3}ight) \

& =\frac{-1}{(x-3)^{2}}

\end{aligned}

$$

The slope of the tangent is 2

So, we have:

$$

\begin{aligned}

& \frac{-1}{(x-3)^{2}}=2 \

& \Rightarrow 2(x-3)^{2}=-1 \

& \Rightarrow(x-3)^{2}=\frac{-1}{2}

\end{aligned}

$$

It is not possible since the LHS is positive and RHS is negative

Thus, there is no tangent to the curve of the slope 2.

Question 13

Find the equations of all lines having slope 0 which are tangent to the curve $y=\frac{1}{x^{2}-2 x+3}$.

Accepted Answer

The given curve is $y=\frac{1}{x^{2}-2 x+3}$

The slope of the tangents to the given curve is given by,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(\frac{1}{x^{2}-2 x+3}ight) \

& =\frac{-(2 x-2)}{\left(x^{2}-2 x+3ight)^{2}} \

& =\frac{-2(x-1)}{\left(x^{2}-2 x+3ight)^{2}}

\end{aligned}

$$

The slope of the tangent is 0

So, we have:

$$

\begin{aligned}

& \frac{-2(x-1)}{\left(x^{2}-2 x+3ight)^{2}}=0 \

& \Rightarrow-2(x-1)=0 \

& \Rightarrow x=1

\end{aligned}

$$

When, $x=1$

Then,

$$

\begin{aligned}

y & =\frac{1}{1-2+3} \

& =\frac{1}{2}

\end{aligned}

$$

Hence, the equation of the tangent through $\left(1, \frac{1}{2}ight)$ is given by,

$$

\begin{aligned}

& y-\frac{1}{2}=0(x-1) \

& \Rightarrow y-\frac{1}{2}=0 \

& \Rightarrow y=\frac{1}{2}

\end{aligned}

$$

Thus, the equation of the line is $y=\frac{1}{2}$.

Question 14

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are

(i) parallel to $x$-axis

(ii) parallel to $y$-axis

Accepted Answer

The equation of the given curve is $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$

On differentiating both sides with respect to $x$, we have:

$$

\begin{aligned}

& \Rightarrow \frac{2 x}{9}+\frac{2 y}{16} \cdot \frac{d y}{d x}=0 \

& \Rightarrow \frac{d y}{d x}=\frac{-16 x}{9 y}

\end{aligned}

$$

(i) The tangent is parallel to the $x$-axis if the slope of the tangent is $\frac{-16 x}{9 y}=0$, which is possible if $x=0$

Thus,

$$

\frac{x^{2}}{9}+\frac{y^{2}}{16}=1 	ext { for } x=0

$$

Therefore,

$$

\begin{aligned}

& \Rightarrow \frac{y^{2}}{16}=1 \

& \Rightarrow y^{2}=16 \

& \Rightarrow y= \pm 4

\end{aligned}

$$

Hence, the points are $(0,4)$ and $(0,-4)$.

(ii) The tangent is parallel to the $y$-axis if the slope of the normal is 0 , which gives

$$

\begin{aligned}

& \frac{-1}{\left(\frac{-16 x}{9 y}ight)}=0 \

& \Rightarrow \frac{9 y}{16 x}=0 \

& \Rightarrow y=0

\end{aligned}

$$

Thus,

$$

\frac{x^{2}}{9}+\frac{y^{2}}{16}=1 	ext { for } y=0

$$

Therefore,

$$

\begin{aligned}

& \Rightarrow \frac{x^{2}}{9}=1 \

& \Rightarrow x^{2}=9 \

& \Rightarrow x= \pm 3

\end{aligned}

$$

Hence, the points are $(3,0)$ and $(-3,0)$.

Question 15

Find the equation of the tangents and normal to the given curves at the indicated points

(i) $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at $(0,5)$

(ii) $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at $(1,3)$

(iii) $y=x^{3}$ at $(1,1)$

(iv) $y=x^{2}$ at $(0,0)$

(v) $x=\cos t, y=\sin t$ at $t=\frac{\pi}{4}$

Accepted Answer

(i) The equation of the curve is $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$

On differentiating with respect to $x$, we get:

$$

\begin{aligned}

& \frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10 \

& \left.\frac{d y}{d x}ight]_{(0,5)}=-10

\end{aligned}

$$

Thus, the slope of the tangent at $(0,5)$ is -10 .

The equation of the tangent is given as:

$$

\begin{aligned}

& y-5=-10(x-0) \

& \Rightarrow y-5=-10 x \

& \Rightarrow 10 x+y=5

\end{aligned}

$$

Slope of normal at $(0,5)$ is

$$

\frac{-1}{	ext { slope of the tangent at }(0,5)}=\frac{-1}{-10}=\frac{1}{10}

$$

Therefore, the equation of the normal at $(0,5)$ is given as:

$$

\begin{aligned}

& y-5=\frac{1}{10}(x-0) \

& \Rightarrow 10 y-50=x \

& \Rightarrow x-10 y+50=0

\end{aligned}

$$

(ii) The equation of the curve is $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at $(1,3)$

On differentiating with respect to $x$, we get:

$$

\begin{aligned}

& \frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10 \

& \left.\frac{d y}{d x}ight]_{(1,3)}=4-18+26-10=2

\end{aligned}

$$

Thus, the slope of the tangent at $(1,3)$ is 2 .

The equation of the tangent is given as:

$$

\begin{aligned}

& y-3=2(x-1) \

& \Rightarrow y-3=2 x-2 \

& \Rightarrow y=2 x+1

\end{aligned}

$$

Slope of normal at $(1,3)$ is

$$

\frac{-1}{	ext { slope of the tangent at }(1,3)}=\frac{-1}{2}

$$

Therefore, the equation of the normal at $(1,3)$ is given as:

$$

\begin{aligned}

& y-3=\frac{1}{2}(x-1) \

& \Rightarrow 2 y-6=x+1 \

& \Rightarrow x+2 y-7=0

\end{aligned}

$$

(iii) The equation of the curve is $y=x^{3}$ at $(1,1)$

On differentiating with respect to $x$, we get:

$$

\begin{aligned}

& \frac{d y}{d x}=3 x^{2} \

& \left.\frac{d y}{d x}ight]_{(1,1)}=3(1)^{2}=3

\end{aligned}

$$

Thus, the slope of the tangent at $(1,1)$ is 3 .

The equation of the tangent is given as:

$$

\begin{aligned}

& y-1=3(x-1) \

& \Rightarrow y=3 x-2

\end{aligned}

$$

Slope of normal at $(1,1)$ is

$$

\frac{-1}{	ext { slope of the tangent at }(1,1)}=\frac{-1}{3}

$$

Therefore, the equation of the normal at $(1,1)$ is given as:

$$

\begin{aligned}

& y-1=\frac{-1}{3}(x-1) \

& \Rightarrow 3 y-3=-x+1 \

& \Rightarrow x+3 y-4=0

\end{aligned}

$$

(iv) The equation of the curve is $y=x^{2}$ at $(0,0)$

On differentiating with respect to $x$, we get:

$$

\begin{aligned}

& \frac{d y}{d x}=2 x \

& \left.\frac{d y}{d x}ight]_{(0,0)}=0

\end{aligned}

$$

Thus, the slope of the tangent at $(0,0)$ is 0 .

The equation of the tangent is given as:

$$

\begin{aligned}

& y-0=0(x-0) \

& \Rightarrow y=0

\end{aligned}

$$

Slope of normal at $(0,0)$ is

$$

\frac{-1}{	ext { slope of the tangent at }(0,0)}=\frac{-1}{0} 	ext {, which is not defined. }

$$

Therefore, the equation of the normal at $(0,0)$ is given as $x=0$

The equation of the curve is $x=\cos t$ and $y=\sin t$ at $t=\frac{\pi}{4}$

On differentiating with respect to $t$, we get:

$$

\frac{d x}{d t}=-\sin t \quad 	ext { and } \quad \frac{d y}{d t}=\cos t

$$

Therefore,

$$

\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}ight)}{\left(\frac{d x}{d t}ight)}=\frac{\cos t}{-\sin t}=-\cot t

$$

Hence,

$$

\left.\frac{d y}{d x}ight]_{t=\frac{\pi}{4}}=-\cot t=-1

$$

Thus, the slope of the tangent at $t=\frac{\pi}{4}$ is -1 .

Hence,

$$

x=\frac{1}{\sqrt{2}} 	ext { and } y=\frac{1}{\sqrt{2}}

$$

The equation of the tangent is given as:

$$

\begin{aligned}

& \Rightarrow y-\frac{1}{\sqrt{2}}=-1\left(x-\frac{1}{\sqrt{2}}ight) \

& \Rightarrow x+y-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0 \

& \Rightarrow x+y-\sqrt{2}=0

\end{aligned}

$$

Slope of normal at $t=\frac{\pi}{4}$ is

$$

\frac{-1}{	ext { slope of the tangent at } t=\frac{\pi}{4}}=\frac{-1}{-1}=1

$$

Therefore, the equation of the normal at $t=\frac{\pi}{4}$ is given as:

$$

\begin{aligned}

& y-\frac{1}{\sqrt{2}}=1\left(x-\frac{1}{\sqrt{2}}ight) \

& \Rightarrow x=y

\end{aligned}

$$

Question 16

Find the equation of the tangent line to the curve $y=x^{2}-2 x+7$ which is

(a) parallel to the line $2 x-y+9=0$

(b) perpendicular to the line $5 y-15 x=13$

Accepted Answer

The equation of the curve is $y=x^{2}-2 x+7$

On differentiating with respect to $x$, we get:

$$

\frac{d y}{d x}=2 x-2

$$

(a) The equation of the line is $2 x-y+9=0$

$$

\Rightarrow y=2 x+9

$$

This is of the form $y=m x+c$

Hence, the slope of line is 2

If a tangent is parallel to the line $2 x-y+9=0$, then the slope of the tangent is equal to the slope of the line.

Therefore, we have:

$$

\begin{aligned}

& 2=2 x-2 \

& \Rightarrow 2 x=4 \

& \Rightarrow x=2

\end{aligned}

$$

Now, $x=2$

Then,

$$

\begin{aligned}

& \Rightarrow y=4-4+7 \

& \Rightarrow y=7

\end{aligned}

$$

Thus, the equation of tangent passing through $(2,7)$ is given by,

$$

\begin{aligned}

& y-7=2(x-2) \

& \Rightarrow y-2 x-3=0

\end{aligned}

$$

(b) perpendicular to the line $5 y-15 x=13$

$$

\Rightarrow y=3 x+\frac{13}{5}

$$

This is of the form $y=m x+c$

Hence, the slope of line is 3

If a tangent is perpendicular to the line $5 y-15 x=13$, then the slope of the tangent is,

$$

\frac{-1}{	ext { slope of the line }}=\frac{-1}{3}

$$

Therefore, we have:

$$

\begin{aligned}

& 2 x-2=\frac{-1}{3} \

& \Rightarrow 2 x=\frac{-1}{3}+2 \

& \Rightarrow 2 x=\frac{5}{3} \

& \Rightarrow x=\frac{5}{6}

\end{aligned}

$$

Now, $x=\frac{5}{6}$

Then,

$$

\begin{aligned}

y & =\frac{25}{36}+\frac{10}{6}+7 \

& =\frac{25-60+252}{36} \

& =\frac{217}{36}

\end{aligned}

$$

Thus, the equation of tangent passing through $\left(\frac{5}{6}, \frac{217}{36}ight)$ is given by,

$$

\begin{aligned}

& y-\frac{217}{36}=\frac{1}{3}\left(x-\frac{5}{6}ight) \

& \Rightarrow \frac{36 y-217}{36}=\frac{-1}{18}(6 x-5) \

& \Rightarrow 36 y-217=-2(6 x-5) \

& \Rightarrow 36 y-217=-12 x+10 \

& \Rightarrow 36 y+12 x-227=0

\end{aligned}

$$

Question 17

Show that the tangents to the curve $y=7 x^{3}+11$ at the points where $x=2$ and $x=-2$ are parallel.

Accepted Answer

The equation of the given curve is $y=7 x^{3}+11$.

Therefore,

$$

\frac{d y}{d x}=21 x^{2}

$$

Thus, the slope of the tangent at the point where $x=2$ is given by,

$$

\left.\frac{d y}{d x}\right]_{x=2}=21(2)^{2}=84

$$

Also, the slope of the tangent at the point where $x=-2$ is given by,

$$

\left.\frac{d y}{d x}\right]_{x=-2}=21(-2)^{2}=84

$$

It is observed clearly that the slopes of the tangents at the points where $x=2$ and $x=-2$ are equal.

Hence, the two tangents are parallel.

Question 18

Find the points on the curve $y=x^{3}$ at which the slope of the tangent is equal to the $y$-coordinates of the point.

Accepted Answer

The equation of the given curve is $y=x^{3}$

Therefore,

$$

\frac{d y}{d x}=3 x^{2}

$$

When the slope of the tangent is equal to the $y$-coordinate of the point, then According to the question, $y=3 x^{2}$

Also, we have $y=x^{3}$

Therefore,

$$

\begin{aligned}

& 3 x^{2}=x^{3} \

& \Rightarrow x^{2}(x-3)=0 \

& \Rightarrow x=0, x=3

\end{aligned}

$$

When, $x=0, \Rightarrow y=3$ and $x=3, \Rightarrow y=3(3)^{2}=27$

Thus, the points are $(0,0)$ and $(3,27)$.

Question 19

For the curve $y=4 x^{3}-2 x^{5}$, find all the points at which the tangent passes through the origin.

Accepted Answer

The equation of the given curve is $y=4 x^{3}-2 x^{5}$

Therefore,

$$

\frac{d y}{d x}=12 x^{2}-10 x^{4}

$$

Hence, the slope of the tangent at the point $(x, y)$ is $12 x^{2}-10 x^{4}$

Thus, the equation of the tangent at $(x, y)$ is given by,

$$

Y-y=\left(12 x^{2}-10 x^{4}ight)(X-x)

$$

When the tangent passes through the origin $(0,0), X=Y=0$.

Therefore,

$$

\begin{aligned}

& -y=\left(12 x^{2}-10 x^{4}ight)(-x) \

& y=12 x^{3}-10 x^{5}

\end{aligned}

$$

Also, we have $y=4 x^{3}-2 x^{5}$

Hence,

$$

\begin{aligned}

& 	herefore 12 x^{3}-10 x^{5}=4 x^{3}-2 x^{5} \

& \Rightarrow 8 x^{5}-8 x^{3}=0 \

& \Rightarrow x^{5}-x^{3}=0 \

& \Rightarrow x^{3}\left(x^{2}-1ight)=0 \

& \Rightarrow x=0, \pm 1

\end{aligned}

$$

When, $x=0, \Rightarrow y=4(0)^{3}-2(0)^{5}=0$

When, $x=1, \Rightarrow y=4(1)^{3}-2(1)^{5}=2$

When, $x=-1, \Rightarrow y=4(-1)^{3}-2(-1)^{5}=-2$

Thus, the points are $(0,0),(1,2)$ and $(-1,-2)$

Question 20

Find the points on the curve $x^{2}+y^{2}-2 x-3=0$ at which the tangents are parallel to the $x$-axis.

Accepted Answer

The equation of the given curve is $x^{2}+y^{2}-2 x-3=0$

On differentiating with respect to $x$, we have:

$$

\begin{aligned}

& 2 x+2 y \frac{d y}{d x}-2=0 \

& \Rightarrow y \frac{d y}{d x}=1-x \

& \Rightarrow \frac{d y}{d x}=\frac{1-x}{y}

\end{aligned}

$$

Now, the tangents are parallel to the $x$-axis if the slope of the tangent is 0 .

Therefore,

$$

\begin{aligned}

& \frac{1-x}{y}=0 \

& \Rightarrow 1-x=0 \

& \Rightarrow x=1

\end{aligned}

$$

But we have $x^{2}+y^{2}-2 x-3=0$ for $x=1$

Hence,

$$

\begin{aligned}

& \Rightarrow y^{2}=4 \

& \Rightarrow y= \pm 2

\end{aligned}

$$

Thus, the points are $(1,2)$ and $(1,-2)$.

Question 21

Find the equation of the normal at the point $\left(a m^{2}, a m^{3}\right)$ for the curve $a y^{2}=x^{3}$.

Accepted Answer

The equation of the given curve is $a y^{2}=x^{3}$

On differentiating with respect to $x$, we have:

$$

\begin{aligned}

& 2 a y \frac{d y}{d x}=3 x^{2} \

& \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{2 a y}

\end{aligned}

$$

The slope of a tangent to the curve at $\left(a m^{2}, a m^{3}ight)$ is

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{\left(a m^{2}, a m^{3}ight)} & =\frac{3\left(a m^{2}ight)^{2}}{2 a\left(a m^{3}ight)} \

& =\frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} \

& =\frac{3 m}{2}

\end{aligned}

$$

Therefore, the slope of normal at $\left(a m^{2}, a m^{3}ight)$ is

$$

\frac{-1}{	ext { slope of the tangent at }\left(a m^{2}, a m^{3}ight)}=\frac{-1}{\left(\frac{3 m}{2}ight)}=\frac{-2}{3 m}

$$

Hence, the equation of the normal at $\left(a m^{2}, a m^{3}ight)$ is given by,

$$

\begin{aligned}

& y-a m^{3}=\frac{-2}{3 m}\left(x-a m^{2}ight) \

& \Rightarrow 3 m y-3 a m^{4}=-2 x+2 a m^{2} \

& \Rightarrow 2 x+3 m y-a m^{2}\left(2+3 m^{2}ight)=0

\end{aligned}

$$

Question 22

Find the equation of the normal to the curve $y=x^{3}+2 x+6$ which are parallel to the line $x+14 y+4=0$.

Accepted Answer

The equation of the given curve is $y=x^{3}+2 x+6$

The slope of the tangent to the given curve at any point $(x, y)$ is given by,

$$

\frac{d y}{d x}=3 x^{2}+2

$$

Therefore, slope of the normal to the given curve is,

$$

\frac{-1}{	ext { slope of the tangent }}=\frac{-1}{3 x^{2}+2}

$$

The equation of the given line is $x+14 y+4=0$

$$

\Rightarrow y=-\frac{1}{14} x-\frac{4}{14}, 	ext { which is the form of } y=m x+c

$$

Hence,

$$

\begin{aligned}

& \frac{-1}{3 x^{2}+2}=\frac{-1}{14} \

& \Rightarrow 3 x^{2}+2=14 \

& \Rightarrow 3 x^{2}=12 \

& \Rightarrow x^{2}=4 \

& \Rightarrow x= \pm 2

\end{aligned}

$$

When, $x=2, \Rightarrow y=8+4+6=18$

When, $x=-2, \Rightarrow y=-8-4+6=-6$

Therefore, there are two normal to the given curve with slope $\frac{-1}{14}$ and passing through the points $(2,18)$ and $(-2,-6)$.

Thus, the equation of the normal through $(2,18)$ is

$$

\begin{aligned}

& y-18=\frac{-1}{14}(x-2) \

& \Rightarrow 14 y-252=x+2 \

& \Rightarrow x+14 y-254=0

\end{aligned}

$$

And the equation of the normal through $(-2,-6)$ is

$$

\begin{aligned}

& y-(-6)=\frac{-1}{14}(x-(-2)) \

& \Rightarrow 14 y+84=-x-2 \

& \Rightarrow x+14 y+86=0

\end{aligned}

$$

Hence, the equations of the normal to the given curve are $x+14 y-254=0$ and $x+14 y+86=0$.

Question 23

Find the equation of the tangent and normal to the parabola $y^{2}=4 a x$ at the point $\left(a t^{2}, 2 a t\right)$.

Accepted Answer

The equation of the given parabola is $y^{2}=4 a x$.

On differentiating both sides with respect to $x$, we have:

$$

\begin{aligned}

& 2 y \frac{d y}{d x}=4 a \

& \Rightarrow \frac{d y}{d x}=\frac{2 a}{y}

\end{aligned}

$$

Therefore, the slope of the tangent at $\left(a t^{2}, 2 a tight)$ is

$$

\left.\frac{d y}{d x}ight]_{\left(a t^{2}, 2 a tight)}=\frac{2 a}{2 a t}=\frac{1}{t}

$$

Hence, the equation of the tangent at $\left(a t^{2}, 2 a tight)$ is

$$

\begin{aligned}

& y-2 a t=\frac{1}{t}\left(x-a t^{2}ight) \

& \Rightarrow t y-2 a t^{2}=x-a t^{2} \

& \Rightarrow t y=x+a t^{2}

\end{aligned}

$$

Now, the slope of the normal at $\left(a t^{2}, 2 a tight)$ is

$$

\frac{-1}{	ext { slope of the tangent at }\left(a t^{2}, 2 a tight)}=\frac{-1}{\left(\frac{1}{t}ight)}=-t

$$

Thus, the equation of the normal at $\left(a t^{2}, 2 a tight)$ is

$$

\begin{aligned}

& y-2 a t=-t\left(x-a t^{2}ight) \

& \Rightarrow y-2 a t=-t x+a t^{3} \

& \Rightarrow y=-t x+2 a t+a t^{3}

\end{aligned}

$$

Question 24

Prove that the curves $x=y^{2}$ and $x y=k$ cut at right angles if $8 k^{2}=1$.

[Hint: Two curves intersect at right angle if the tangents to the curve at the point of intersection are perpendicular to each other.]

Accepted Answer

The equations of the given curves are $x=y^{2}$ and $x y=k$

Putting $x=y^{2}$ in $x y=k$

$$

\begin{aligned}

& y^{3}=k \

& \Rightarrow y=k^{\frac{1}{3}} \

& \Rightarrow x=k^{\frac{2}{3}}

\end{aligned}

$$

Thus, the point of intersection of the given curves is $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)$.

Differentiating $x=y^{2}$ with respect to $x$, we have:

$$

\begin{aligned}

& 1=2 y \frac{d y}{d x} \

& \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}

\end{aligned}

$$

Therefore, the slope of the tangent to the curve $x=y^{2}$ at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)$ is

$$

\left.\frac{d y}{d x}ight]_{\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)}=\frac{1}{2 k^{\frac{1}{3}}}

$$

On differentiating $x y=k$ with respect to $x$, we have:

$$

\begin{aligned}

& x \frac{d y}{d x}+y=0 \

& \Rightarrow \frac{d y}{d x}=\frac{-y}{x}

\end{aligned}

$$

Now, slope of the tangent to the curve $x y=k$ at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)$ is

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)} & \left.=\frac{-y}{x}ight]_{\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)} \

& =-\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}} \

& =\frac{-1}{k^{\frac{1}{3}}}

\end{aligned}

$$

We know that two curves intersect at right angles if the tangents to the curves at the point of intersection i.e., at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)$ are perpendicular to each other.

This implies that we should have the product of the tangents as -1 .

Thus, the given two curves cut at right angles if the product of the slopes of their respective tangents at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}ight)$ is -1 . i.e.,

$$

\begin{aligned}

& \left(\frac{1}{2 k^{\frac{1}{3}}}ight)\left(\frac{-1}{k^{\frac{1}{3}}}ight)=-1 \

& \Rightarrow 2 k^{\frac{2}{3}}=1 \

& \Rightarrow\left(2 k^{\frac{2}{3}}ight)^{3}=(1)^{3} \

& \Rightarrow 8 k^{2}=1

\end{aligned}

$$

Hence, the given curves cut at right angle if $8 k^{2}=1$.

Question 25

Find the equation of the tangent and normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the point $\left(x_{0}, y_{0}\right)$.

Accepted Answer

Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$, we have:

$$

\begin{aligned}

& \frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0 \

& \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{2 x}{a^{2}} \

& \Rightarrow \frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}

\end{aligned}

$$

Therefore, the slope of the tangent at $\left(x_{0}, y_{0}ight)$ is

$$

\left.\frac{d y}{d x}ight]_{\left(x_{0}, y_{0}ight)}=\frac{b^{2} x_{0}}{a^{2} y_{0}}

$$

Hence, the equation of tangent at $\left(x_{0}, y_{0}ight)$ is

$$

\begin{aligned}

& y-y_{0}=\frac{b^{2} x_{0}}{a^{2} y_{0}}\left(x-x_{0}ight) \

& \Rightarrow a^{2} y y_{0}-a^{2} y_{0}^{2}=b^{2} x x_{0}-b^{2} x_{0}^{2} \

& \Rightarrow b^{2} x x_{0}-a^{2} y y_{0}-b^{2} x_{0}^{2}-a^{2} y_{0}^{2}=0 \

& \Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}-\left(\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}ight)=0 \

& \Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}-1=0 \

& \Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1

\end{aligned}

$$

Now, the slope of normal at $\left(x_{0}, y_{0}ight)$ is

$$

\frac{-1}{	ext { slope of the tangent at }\left(x_{0}, y_{0}ight)}=\frac{-a^{2} y_{0}}{b^{2} x_{0}}

$$

Hence, the equation of normal at $\left(x_{0}, y_{0}ight)$ is

$$

\begin{aligned}

& y-y_{0}=\frac{-a^{2} y_{0}}{b^{2} x_{0}}\left(x-x_{0}ight) \

& \Rightarrow \frac{y-y_{0}}{a^{2} y_{0}}=\frac{-\left(x-x_{0}ight)}{b^{2} x_{0}} \

& \Rightarrow \frac{y-y_{0}}{a^{2} y_{0}}-\frac{\left(x-x_{0}ight)}{b^{2} x_{0}}=0

\end{aligned}

$$

Question 26

Find the equation of the tangent to the curve $y=\sqrt{3 x-2}$ which is parallel to the line $4 x-2 y+5=0$.

Accepted Answer

The equation of the given curve is $y=\sqrt{3 x-2}$

The slope of the tangent to the given curve at any point $(x, y)$ is given by,

$$

\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}

$$

The equation of the given line is $4 x-2 y+5=0$.

$$

\Rightarrow y=2 x+\frac{5}{2}, 	ext { which is the form of } y=m x+c

$$

Hence, the slope of line is 2 .

Now, the tangent to the given curve is parallel to the line $4 x-2 y+5=0$ if the slope of the tangent is equal to the slope of the line.

Therefore,

$$

\begin{aligned}

& \frac{3}{2 \sqrt{3 x-2}}=2 \

& \Rightarrow \sqrt{3 x-2}=\frac{3}{4} \

& \Rightarrow 3 x-2=\frac{9}{16} \

& \Rightarrow 3 x=\frac{9}{16}+2=\frac{41}{16} \

& \Rightarrow x=\frac{41}{48}

\end{aligned}

$$

When, $x=\frac{41}{48}$

Then,

$$

\begin{aligned}

y & =\sqrt{3\left(\frac{41}{48}ight)-2} \

& =\sqrt{\frac{41}{16}-2} \

& =\sqrt{\frac{41-32}{16}} \

& =\sqrt{\frac{9}{16}} \

& =\frac{3}{4}

\end{aligned}

$$

Thus, equation of the tangent passing through the point $\left(\frac{41}{48}, \frac{3}{4}ight)$ is

$$

\begin{aligned}

& y-\frac{3}{4}=2\left(x-\frac{41}{48}ight) \

& \Rightarrow \frac{4 y-3}{4}=2\left(\frac{48 x-41}{48}ight) \

& \Rightarrow 4 y-3=\left(\frac{48 x-41}{6}ight) \

& \Rightarrow 24 y-18=48 x-41 \

& \Rightarrow 48 x-24 y=23

\end{aligned}

$$

Question 27

The line $y=x+1$ is a tangent to the curve $y^{2}=4 x$ at the point

(A) $(1,2)$

(B) $(2,1)$

(C) $(1,-2)$

(D) $(-1,2)$

Accepted Answer

The equation of the given curve is $y^{2}=4 x$.

Differentiating with respect to $x$, we have:

$$

\begin{aligned}

& 2 y \frac{d y}{d x}=4 \

& \Rightarrow \frac{d y}{d x}=\frac{2}{y}

\end{aligned}

$$

The given line is $y=x+1$ which is of the form $y=m x+c$.

Hence, slope of the line is 1

The line $y=x+1$ is a tangent to the given curve if the slope of the line is equal to the slope of the tangent.

Also, the line must intersect the curve.

Thus, we must have:

$$

\begin{aligned}

& \frac{2}{y}=1 \

& \Rightarrow y=2

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

& y=x+1 \

& \Rightarrow x=y-1 \

& \Rightarrow x=2-1 \

& \Rightarrow x=1

\end{aligned}

$$

Hence, the line $y=x+1$ is a tangent to the given curve at the point $(1,2)$.

Thus, the correct answer is $\mathbf{A}$.

\section*{EXERCISE 6.4}