Ex 6.3 class 12 maths ncert solutions

The given curve \( y=3 x^{4}-4 x \)
Then, the slope of the tangent to the given curve at \( x=4 \) is given by,
\(
\left.\left.\frac{d y}{d x}\right]_{x=4}=12 x^{3}-4\right]_{x=4}=12(4)^{3}-4\)
\(=12(64)-4\)
\(=764
\)
Therefore, the slop of the tangent is 764.

The given curve is \( y=\frac{x-1}{x-2} \)
\(
\therefore \frac{d y}{d x}=\frac{(x-2)(1)-(x-1)(1)}{(x-2)^{2}}\)
\(=\frac{x-2-x+1}{(x-2)^{2}}\)
\(=\frac{-1}{(x-2)^{2}}
\)
Then, the slope of the tangent
\(
\left.\left.\frac{d y}{d x}\right]_{x=10}=\frac{-1}{(x-2)^{2}}\right]_{x=10}=\frac{-1}{(10-2)^{2}}=\frac{-1}{64}
\)
Therefore, the slope of the tangent is \( \frac{-1}{64} \).

The given curve is \( y=x^{3}-x+1 \)
\(
\therefore \frac{d y}{d x}=3 x^{2}-1
\)
Then, the slope of the tangent
\(
\left.\left.\frac{d y}{d x}\right]_{x=4}=3 x^{2}-1\right]_{x=2}=3(2)^{2}-1\)
\(=12-1=11
\)
Therefore, the slope of the tangent 11 .

The given curve is \( y=x^{3}-3 x+2 \)
\(
\therefore \frac{d y}{d x}=3 x^{2}-3
\)
Then, the slope of the tangent
\(
\left.\left.\frac{d y}{d x}\right]_{x=3}=3 x^{2}-3\right]_{x=3}=3(3)^{2}-3
\)
\(
=27-3=24
\)
Therefore, the slope of the tangent 24 .

The given curve is \( y=\operatorname{acos}^{3} \theta \) and \( y=\operatorname{asin}^{3} \theta \)
\(
\therefore \frac{d x}{d \theta}=3 \operatorname{acos}^{2} \theta(-\sin \theta)=-3 a \cos ^{2} \theta \sin \theta\)
\(\frac{d y}{d \theta}=3 \mathrm{asi} n^{2} \theta \cos \theta\)
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a sin^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}\)
\(=-\frac{\sin \theta}{\cos \theta}=-\tan \theta
\)
Then, the slope of the tangent \( \theta=\frac{\pi}{4} \) is given by:
\(
\left.\left.\frac{d y}{d x}\right]_{\theta=\frac{\pi}{4}}=-\tan \theta\right]_{\theta=\frac{\pi}{4}}=-\tan \frac{\pi}{4}=-1
\)
Then, the slope of the tangent \( \theta=\frac{\pi}{4} \) is given by: \( \frac{1}{\text { slope of the tangent at } \theta=\frac{\pi}{4}}=\frac{-1}{-1}=1 \)
Therefore, the slope of the tangent 1 .

The given curve is \( x=1-a \sin \theta \) and \( y=b \cos ^{2} \theta \)
\(
\therefore \frac{d x}{d \theta}=-\mathrm{a} \cos \theta\)
\(\frac{d y}{d \theta}=-2 \mathrm{~b} \sin \theta \cos \theta\)
\(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-2 \mathrm{~b} \sin \theta \cos \theta}{-a \cos \theta}=\frac{2 b}{a} \sin \theta
\)
Then, the slope of the tangent \( \theta=\frac{\pi}{2} \) is given by:
\(
\left.\left.\frac{d y}{d x}\right]_{\theta=\frac{\pi}{2}}=\frac{2 b}{a} \sin \theta\right]_{\theta=\frac{\pi}{2}}=\frac{2 b}{a} \sin \frac{\pi}{2}=\frac{2 b}{a}
\)
Then, the slope of the tangent \( \theta=\frac{\pi}{2} \) is given by:
\(
\frac{1}{\text { slope of the tangent at } \theta=\frac{\pi}{2}}=\frac{-1}{\left(\frac{2 b}{a}\right)}=-\frac{a}{2 b}
\)
Therefore, the slope of the tangent \( -\frac{a}{2 b} \).

It is given that the curve \( y=x^{3}-3 x^{2}-9 x+7 \)
\(
\frac{d y}{d x}=3 x^{2}-6 x-9
\)
We know that the tangent is parallel to the \(x\)-axis if the slope of the tangent is zero.
Therefore, \( 3 x^{2}-6 x-9=0 \)
\(
\Rightarrow x^{2}-2 x-3=0\)
\(\Rightarrow(x-3)(x+1)=0\)
\(\Rightarrow x=3 \text { and } x=-1
\)
When \( x=3, y=(3)^{3}-3(3)^{2}-9(3)+7=27-27-27+7=-20 \)
When \( x=-1, y=(-1)^{3}-3(-1)^{2}-9(-1)+7=-1-3+9+7=12 \)
Therefore, the points at which the tangent is parallel to the \(x\) - axis are \((3, -20)\) and \( (-1,12) \)

We know that if a tangent is parallel to the chord joining the points \( (2,0) \) and \( (4,4) \), then
Slope of the tangent \( = \) slope of the curve \(\ldots(1)\)
And, the slope of the curve \( =\frac{4-0}{4-2}=2 \)
Now, slope of the tangent to the given curve at a point \( (x, y) \) is:
\(
\frac{d y}{d x}=2(x-2)
\)
Now, from (1) we have,
\(
2(x-2)=2\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3
\)
So, when \( x=3 \) then \( y=(3-2)^{2}=1 \)
Therefore, required points are \( (3,1) \).

It is given that equation of the curve \( y=x^{3}-11 x+5 \)
At which the tangent is \( y=x-11 \)
\( \Rightarrow \) Slope of the tangent \( =1 \)
Now, slope of the tangent to the given curve at a point \( (x, y) \) is:
\(
\frac{d y}{d x}=3 x^{2}-11\)
\(\Rightarrow 3 x^{2}-11=1\)
\(\Rightarrow 3 x^{2}=12\)
\(\Rightarrow x^{2}=4\)
\(\Rightarrow x= \pm 2
\)
So, when \( x=2 \) then \( y=(2)^{3}-11(2)+5=-9 \)
And when \( x=-2 \) then \( y=(-2)^{3}-11(-2)+5=19 \)
Therefore, required points are \( (2,-9) \) and \( (-2,19) \).

It is given that equation of the curve \( y=\frac{1}{x-1}, x \neq 1 \)
Now, slope of the tangent to the given curve at a point \( (x, y) \) is:
\(
\frac{d y}{d x}=\frac{-1}{(x-1)^{2}}
\)
Now, if the slope of the tangent is -1 , then we get,
\(
\frac{-1}{(x-1)^{2}}=-1\)
\(\Rightarrow(x-1)^{2}=1\)
\(\Rightarrow(x-1)= \pm 1\)
\(\Rightarrow x=2,0
\)
So, when \( x=2 \) then \( y=1 \)
And when \( x=0 \) then \( y=1 \)
Therefore, required points are \( (0,-1) \) and \( (2,1) \).
Now, the equation of the tangent \( (0,1) \) is given by:
\(
y-(-1)=-1(x-0)\)
\(\Rightarrow y+1=-x\)
\(\Rightarrow y+x+1=0
\)
And the equation of the tangent \( (2,1) \) is given by:
\(
y-1=-1(x-2)\)
\(\Rightarrow y-1=-x+2
\)
\(
\Rightarrow y+x-3=0
\)
Therefore, the equations of the required lines are \( y+x+1=0 \) and \( y+x \) \( -3=0 \).

It is given that equation of the curve \( y=\frac{1}{x-3}, x \neq 3 \)
Now, slope of the tangent to the given curve at a point \( (x, y) \) is:
\(
\frac{d y}{d x}=\frac{-1}{(x-3)^{2}}
\)
Now, if the slope of the tangent is 2 , then we get,
\(
\frac{-1}{(x-3)^{2}}=2\)
\(\Rightarrow 2(x-3)^{2}=-1\)
\(\Rightarrow(x-3)^{2}=\frac{-1}{2}
\)
This is not possible since the L.H.S. is positive while the R.H.S. is negative.
Therefore, there is no tangent to the given curve having a slope 2 .

It is given that equation of the curve \( y=\frac{1}{x^{2}-2 x+3} \),
Now, slope of the tangent to the given curve at a point \( (x, y) \) is:
\(
\frac{d y}{d x}=\frac{-(2 x-2)}{\left(x^{2}-2 x+3\right)^{2}}=\frac{-2(x-1)}{\left(x^{2}-2 x+3\right)^{2}}
\)
Now, if the slope of the tangent is 0 , then we get,
\(
\frac{-2(x-1)}{\left(x^{2}-2 x+3\right)^{2}}=0\)
\(\Rightarrow-2(x-1)=0\)
\(\Rightarrow x=1
\)
So, when \( x=1 \) then \( y=\frac{1}{1-2+3}=\frac{1}{2} \)
Now, the equation of the tangent \( \left(0, \frac{1}{2}\right) \) is given by:
\(
y-\frac{1}{2}=0(x-1)\)
\(\Rightarrow y-\frac{1}{2}=0\)
\(\Rightarrow y=\frac{1}{2}
\)
Therefore, the equations of the required line is \( y=\frac{1}{2} \).

It is given that \( \frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \)
Now, differentiating both sides with respect to \(x \), we get \( \frac{2 x}{9}+\frac{2 y}{16} \cdot \frac{d y}{d x}=0 \)
\(
\Rightarrow \frac{d y}{d x}=\frac{-16 x}{9 y}
\)
We know that the tangent is parallel to the \( x \)-axis if the slope is 0 ie, \( \frac{-16 x}{9 y}=0 \), which is possible if \( x=0 \)
Then, \( \frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \quad \) for \( x=0 \)
\(
\begin{array}{l}
\Rightarrow y^{2}=16 \\
\Rightarrow y= \pm 4
\end{array}
\)
Therefore, the points at which the tangents are parallel to the \(x\) -axis are \( (0,4) \) and \( (0,-4) \).

It is given that \( \frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \)
Now, differentiating both sides with respect to \(x\) , we get
\(
\frac{2 x}{9}+\frac{2 y}{16} \cdot \frac{d y}{d x}=0\)
\(\Rightarrow \frac{d y}{d x}=\frac{-16 x}{9 y}
\)
We know that the tangent is parallel to the \(y\)-axis if the slope of the normal is 0 ie,
\(
\frac{1}{\frac{-16 x}{9 y}}=\frac{9 y}{16 x}=0\)
\(\Rightarrow y=0
\)
Then, \( \frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \) for \( y=0 \)
\(
\Rightarrow x= \pm 3
\)
Therefore, the points at which the tangents are parallel to the \(y\) -axis are \( (3,0) \) and \( (-3,0) \).

It is given that equation of curve is \( y=x^{4}-6 x^{3}+13 x^{2}-10 x+5 \)
On differentiating with respect to \( x \), we get
\(
\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10
\)
Now, Slope of tangent will be the value of \( \frac{d y}{d x} \) at \( x=0 \) i.e.
\(
m_{1}=4(0)^{3}-18(0)^{2}+26(0)-10=-10
\)
Therefore, the slope of the tangent at \( (0,5) \) is -10 .
Then, the equation of the tangent is
\(
y-5=-10(x-0)\)
\(\Rightarrow y-5=10 x\)
\(\Rightarrow 10 x+y=5
\)
Also, slope of normal at \( (0,5) \)
\(
=\frac{-1}{\text { Slope of the tangent at }(0,5)}=\frac{1}{10}
\)
Now, equation of the normal at \( (0,5) \)
\(
y-5=\frac{1}{10}(x-0)\)
\(\Rightarrow 10 y-50=x\)
\(\Rightarrow x-10 y+50=0
\)

It is given that equation of curve is \( y=x^{4}-6 x^{3}+13 x^{2}-10 x+5 \)
On differentiating with respect to \(x ,\) we get
\(
\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10\)
\(\left.\frac{d y}{d x}\right]_{(1,3)}=4-18+26-10=2
\)
Therefore, the slope of the tangent at \( (1,3) \) is 2 .
Then, the equation of the tangent is
\(
y-3=2(x-1)\)
\(\Rightarrow y-3=2 x-2\)
\(\Rightarrow y=2 x+1
\)
Then, slope of normal at \( (1,3) \)
\(
=\frac{-1}{\text { Slope of the tangent at }(1,3)}=\frac{-1}{2}
\)
Now, equation of the normal at \( (1,3) \)
\(
y-3=-\frac{1}{2}(x-1)\)
\(\Rightarrow 2 y-6=-x+1\)
\(\Rightarrow x+2 y-7=0
\)

It is given that equation of curve is \( y=x^{3} \)
On differentiating with respect to \(x \), we get
\(\frac{d y}{d x}=3 x^{2}\)
\(\left.\frac{d y}{d x}\right]_{(1,1)}=3\)
Therefore, the slope of the tangent at \( (1,1) \) is 3 .
Then, the equation of the tangent is
\(y-1=3(x-1)\)
\(\Rightarrow y=3 x-2\)
Then, slope of normal at \( (1,1) \)
\(=\frac{-1}{\text { Slope of the tangent at }(1,1)}=\frac{-1}{3}\)
Now, equation of the normal at \( (1,1) \)
\(y-1=\frac{-1}{3}(x-1)\)
\(\Rightarrow 3 y-3=-x+1\)
\(\Rightarrow x+3 y-4=0\)

It is given that equation of curve is \( y=x^{2} \)
On differentiating with respect to \(x \), we get
\(\frac{d y}{d x}=2 x\)
\(\left.\frac{d y}{d x}\right]_{(0,0)}=0\)
Therefore, the slope of the tangent at \( (0,0) \) is 0
Then, the equation of the tangent is
\(y-0=0(x-0)\)
\(\Rightarrow y=0\)
Then, slope of normal at \( (0,0) \)
\( =\frac{-1}{\text { Slope of the tangent at }(0,0)}=\frac{-1}{0} \), which is not defined
Now, equation of the normal at \( (0,0) \)
\(x=0\)

It is given that equation of curve is \( x=\cos t, y=\sin t \)
\(\frac{d x}{d t}=-\sin t, \quad \frac{d y}{d t}=\cos t\)
On differentiating with respect to \(x \), we get
\(\frac{d x}{d t}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\cos t}{-\sin t}=-\operatorname{cot}t\)
\(\left.\frac{d y}{d x}\right]_{t=\frac{\pi}{4}}=-\cot \mathrm{t}=-1\)
Therefore, the slope of the tangent at \( t=\frac{\pi}{4} \) ) is -1 .
When \( \mathrm{t}=\frac{\pi}{4}, x=\frac{1}{\sqrt{2}} \) and \( y=\frac{1}{\sqrt{2}} \)
Then, the equation of the tangent is \( t=\frac{\pi}{4} \) is
\(y-\frac{1}{\sqrt{2}}=-1\left(x-\frac{1}{\sqrt{2}}\right)\)
\(\Rightarrow x+y-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\)
\(\Rightarrow x+y-\sqrt{2}=0\)
Then, slope of normal at \( t=\frac{\pi}{4} \)
\(=\frac{-1}{\text { Shape of the tangent at } t=\frac{\pi}{4}}=1\)
Now, equation of the normal at \( \mathrm{t}=\frac{\pi}{4} \)
\(y-\frac{1}{\sqrt{2}}=1\left(x-\frac{1}{\sqrt{2}}\right)\)
\(\Rightarrow x=y\)

It is given that equation of the curve is \( y=x^{2}-2 x+7 \)
On differentiating with respect to \(x \), we get
\(\frac{d y}{d x}=2 x-2\)
The equation of the line is \( 2 x-y+9=0 \)
\(\Rightarrow y=2 x+9\)
\( \Rightarrow \) Slope of the line \( =2 \)
Now we know that if a tangent is parallel to the line \( 2 x-y+9=0 \), then Slope of the tangent \( = \) Slope of the line
\(\Rightarrow 2=2 x-2\)
\(\Rightarrow 2 x=4\)
\(\Rightarrow x=2\)
Now, putting \( x=2 \), we get
\(y=4-4+7=7
\)
Then, the equation of the tangent passing through \( (2,7) \)
\(
\Rightarrow y-7=2(x-2)\)
\(\Rightarrow y-2 x-3=0
\)
Therefore, the equation of the tangent line to the given curve which is parallel to line \( 2 x-y+9=0 \) is \( y-2 x-3=0 \).

On differentiating with respect to \(x \), we get
\(\frac{d y}{d x}=2 x-2\)
The equation of the line is \( 5 y-15 x=13 \)
\(\Rightarrow y=3 x+\frac{13}{5}\)
\( \Rightarrow \) Slope of the line \( =3 \)
Now we know that if a tangent is perpendicular to the line \( 5 y-15 x= \) 13 , then
\(\frac{-1}{\text { Slope of the line }}=\frac{-1}{3}\)
\(\Rightarrow 2 x-2=\frac{-1}{3}\)
\(\Rightarrow 2 x=\frac{-1}{3}+2\)
\(\Rightarrow x=\frac{5}{6}\)
Now, putting \( x=\frac{5}{6} \), we get
\(y=\frac{25}{36}-\frac{10}{6}+7=\frac{217}{36}\)
Then, the equation of the tangent passing through \( \left(\frac{5}{6}, \frac{217}{36}\right) \)
\(\Rightarrow y-\frac{217}{36}=\frac{-1}{3}\left(x-\frac{5}{6}\right)\)
\(\Rightarrow \frac{36 y-217}{36}=\frac{-1}{18}(6 x-5)\)
\(\Rightarrow 36 y-217=-2(6 x-5)\)
\(\Rightarrow 36 y+12 x-227=0\)
Therefore, the equation of the tangent line to the given curve which is perpendicular to line \( 5 y-15 x=13 \) is \( 36 y+12 x-227=0 \).

The given curve \( y=7 x^{3}+11 \)
Then, the slope of the tangent to the given curve at \( x=4 \) is given by,
\(\left.\left.\frac{d y}{d x}\right]_{x=2}=21 x^{2}\right]_{x=2}=21(2)^{3}=84\)
It is cleared that the slopes of the tangents at the points where \( x=2 \) and \( x=-2 \) are equal.
Therefore, the two tangents are parallel.

The given curve \( y=x^{3} \)
\(\frac{d y}{d x}=3 x^{2}\)
Then, the slope of the tangent at the point ( \( x, y \) ) is given by:
\(\left.\frac{d y}{d x}\right]_{(x, y)}=3 x^{2}\)
We know that, when the slope of the tangent is equal to the \( y \) - coordinate of the point,
Then \( y=3 x^{2} \)
Also, we have \( y=x^{3} \)
\(\Rightarrow 3 x^{2}=x^{3}\)
\(\Rightarrow x^{2}(x-3)=0\)
\(\Rightarrow x=0, x=3\)
When \( x=0 \) then \( y=0 \)
and when \( x=3 \) then \( y=27 \)
Therefore, the required points are \( (0,0) \) and \( (3,27) \).

The given curve \( y=4 x^{3}-2 x^{5} \)
\(\frac{d y}{d x}=12 x^{2}-10 x^{4}\)
Then, the slope of the tangent at the point \( (x, y) \) is \( 12 x^{2}-10 x^{4} \)
The equation of the tangent at ( \( x, y \) ) is given by,
\(Y-y=\left(12 x^{2}-10 x^{4}\right)(X-x)\ldots(1)\)
When the tangent passes through the origin \( (0,0) \), then \( x=y=0 \)
Then equation (1) becomes,
\(-y=\left(12 x^{2}-10 x^{4}\right)(-x)\)
\(\Rightarrow y=\left(12 x^{3}-10 x^{5}\right)\)
Also, we have \( y=4 x^{3}-2 x^{5} \)
\(\Rightarrow\left(12 x^{3}-10 x^{5}\right)=4 x^{3}-2 x^{5}\)
\(\Rightarrow 8 x^{5}-8 x^{3}=0\)
\(\Rightarrow x^{5}-2 x^{3}=0\)
\(\Rightarrow x^{3}\left(x^{2}-1\right)=0\)
\(\Rightarrow x=0, \pm 1\)
When \( x=0 \) then \( y=0 \)
When \( x=1 \) then \( y=2 \)
And when \( x=-1 \) then \( y=-2 \)
Therefore, the required points are \( (0,0),(1,2) \) and \( (-1,-2) \).

It is given that \( x^{2}+y^{2}-2 x-3=0 \)
Now, differentiating both sides with respect to \(x \), we get
\(2 x+2 y \cdot \frac{d y}{d x}=0\)
\(\Rightarrow y \frac{d y}{d x}=1-x\)
\(\Rightarrow \frac{d y}{d x}=\frac{1-x}{y}\)
We know that the tangents are parallel to the \(x\) -axis if the slope of the tangent is 0 ie,
\(\frac{1-x}{y}=0\)
\(\Rightarrow 1-x=0\)
\(\Rightarrow x=1\)
But, \( x^{2}+y^{2}-2 x-3=0 \) for \( x=1 \)
\(\Rightarrow y^{2}=4\)
\(\Rightarrow y= \pm 2\)
Therefore, the points at which the tangents are parallel to the \(x\) -axis are \( (1,2) \) and \( (1,-2) \).

It is given that \( ay^{2}=x^{3} \)
Now, differentiating both sides with respect to \( x \), we get
\( 2ay. \frac{d y}{d x}=3 x^{2} \)
\( \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{2 a y} \)
Then, the slope of the tangent to the given curve at \( \left(\mathrm{am}^{2}, \mathrm{am}^{3}\right) \) is
\(\left.\frac{d y}{d x}\right]_{\left(a m^{2}, a m^{3}\right)}=\frac{3\left(a m^{2}\right)^{2}}{2 a\left(a m^{3}\right)}=\frac{3 a^{2} m^{4}}{2 a^{2} m^{3}}=\frac{3 m}{2}\)
Then, slope of normal at \( \left(\mathrm{am}^{2}, \mathrm{am}^{3}\right) \)
\(=\frac{-1}{\text { Slope of the tangent at }\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)}=\frac{-2}{3 \mathrm{~m}}\)
Therefore, equation of the normal at \( \left(\mathrm{am}^{2}, \mathrm{am}^{3}\right) \) is given by:
\( y-a m^{3}=\frac{-2}{3 m}\left(x-a m^{2}\right) \)
\( \Rightarrow 3 \mathrm{m}y-3 \mathrm{am}^{4}=-2 x+2 \mathrm{am}^{2} \)
\(\Rightarrow 2 x+3 \mathrm{m}y-\mathrm{am}^{2}\left(2+3 \mathrm{~m}^{2}\right)=0\)
Therefore, equation of the normal at \( \left(\mathrm{am}^{2}, \mathrm{am}^{3}\right) \) is \( 2 x+3 \mathrm{m}y-\mathrm{am}^{2}(2+ \) \( \left.3 \mathrm{~m}^{2}\right)=0 \)

It is given that the equation of the normal to the curve \( y=x^{3}+2 x+6 \) Then, the slope of the tangent to the given curve at any point \( (x, y) \) is given by:
\(\frac{d y}{d x}=3 x^{2}+2\)
Then, slope of normal to the given curve at any point ( \( x, y \) )
\(=\frac{-1}{\text { Slope of thetangent at }(x, y)}=\frac{-1}{3 x^{2}+2}\)
\(=\text { Mediumy }=-\frac{1}{14} x-\frac{4}{14}\)
\( \Rightarrow \) Slope of the given line \( =\frac{-1}{14} \)
We know that if the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.
\(\Rightarrow \frac{-1}{3 x^{2}+2}=\frac{-1}{14}\)
\(\Rightarrow 3 x^{2}+2=14\)
\(\Rightarrow 3 x^{2}=12\)
\(\Rightarrow x^{2}=4\)
\(\Rightarrow x= \pm 2\)
So, when \( x=2 \) then \( y=18 \)
and \( x-2 \) then \( y=-6 \)
Hence, there are two normals to the given curve with the slope \( \frac{-1}{14} \) and passing through the points \( (2,18) \) and \( (-2,-6) \).
Then, the equation of the normal through \( (2,18) \) is:
\(y-18=\frac{-1}{14}(x-2)\)
\(\Rightarrow 14 y-252=-x+2\)
\(\Rightarrow x+14 y-254=0\)
And, the equation of the normal through \( (-2,-6) \) is:
\(y-(-6)=\frac{-1}{14}(x-(-2))\)
\(\Rightarrow y+6=\frac{-1}{14}(x+2)\)
\(\Rightarrow 14 y+84=-x-2\)
\(\Rightarrow x+14 y+86=0\)
Therefore, the equation of the normals to the curve \( y=x^{3}+2 x+6 \) which are parallel to the line \( x+14 y+4=0 \) are \( x+14 y-254=0 \) and \( x \) \( +14 y+86=0 \).

The equation of a parabola is \( y^{2}=4 a x \), then,
On differentiating it with respect to \( x \), we get
\(2 y \frac{d y}{d x}=4 \mathrm{a}\)
\(\Rightarrow \frac{d y}{d x}=\frac{2 a}{y}\)
Then, the slope of the tangent at \( \left(at { }^{2}, 2 \mathrm{at}\right) \) is \( \left.\frac{d y}{d x}\right]_{\left(a t^{2}, 2 a t\right)}=\frac{2 a}{2 a t}=\frac{1}{t} \).
Then, the equation of the tangent at \( \left(\mathrm{at}^{2}, 2 \mathrm{at}\right) \) is given by,
\(y-2 \mathrm{at}=\frac{1}{t}\left(x-a t^{2}\right)\)
\(\Rightarrow \mathrm{ty}-2 a t^{2}=x-a t^{2}\)
\(\Rightarrow \mathrm{ty}-2 \mathrm{at}^{2}=x-\mathrm{at}^{2}\)
\(\Rightarrow \mathrm{ty}=x+\mathrm{at}^{2}\)
Now, Then, slope of normal at \( \left(\mathrm{at}^{2}, 2 \mathrm{at}\right) \)
\(=\frac{-1}{\text { Slope of the tangent at }\left(a t^{2}, 2 a t\right)}=-t\)
Then, the equation of the normal at \( \left(\mathrm{at}^{2}, 2 \mathrm{at}\right) \) is given by:
\(y-2 a t=-t\left(x-a t^{2}\right)\)
\(\Rightarrow y-2 a t=-t x+a t^{3}\)
\(\Rightarrow y=-t x+2 a t+a t^{3}\)
Therefore, the equation of the normal at \( \left(a t^{2}, 2 a t\right) \) is \( y=-t x+2 a t+a t^{3} \).

It is given that the curves \( x=y^{2} \) and \( x y=k \)
Now, putting the value of \( x \) in \( y=k \), we get
\(y^{3}=\mathrm{k}\)
Then, the point of intersection of the given curves is \( \left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right) \)
On differentiating \( x=y^{2} \) with respect to \(x \), we get
\(x \frac{d y}{d x}+y=0\)
\(\Rightarrow \frac{d y}{d x}=\frac{-y}{x}\)
Then, the slope of the tangent at \( xy=\mathrm{k} \) at \( \left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)\)
\(\left.\left.\frac{d y}{d x}\right]_{\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)}=\frac{-y}{x}\right]_{\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)}=\frac{-k^{\frac{1}{3}}}{k^{\frac{2}{3}}}=\frac{-1}{k^{\frac{1}{3}}}\)
As we know that two curves intersect at right angles if the tangents to the curve at the point of intersection are perpendicular to each other.
So, we should have the product of the tangent as -1 .
Then, the given two curves cut at right angles if the product of the slopes of their respective tangent at \( \left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right) \) is -1 .
\(\Rightarrow\left(2 k^{\frac{1}{3}+\frac{1}{3}}\right)=1\)
\(\Rightarrow\left(2 k^{\frac{2}{3}}\right)=1\)
Cubing both side, we get
\(\Rightarrow 8 \mathrm{k}^{2}=1\)
Therefore, the given two curves cut at right angles if \( 8 \mathrm{k}^{2}=1 \).

It is given that the equations of the tangent and normal to the hyperbola \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \)
then,
On differentiating it with respect to \(x \), we get,
\(\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0\)
\(\Rightarrow \frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=\frac{2 x}{a^{2}}\)
\(\Rightarrow \frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}\)
Therefore, the slope of the tangent at \( \left(x_{0}, y_{0}\right) \) is
\(\left.\frac{d y}{d x}\right]_{(x_0, y_0)}=\frac{b^{2} x_0}{a^{2} y_{0}}\)
Then, the equation of the tangent at \( \left(x_{0}, y_{0}\right) \) is given by:
\(y-y_{0}=\frac{b^{2} x}{a^{2} y}\left(x-x_{0}\right)\)
\(\Rightarrow a^{2} y y_{0}-a^{2} y_{0}^{2}=b^{2} x x_{0}-b^{2} x_{0}^{2}\)
\(\Rightarrow b^{2} x x_{0}-a^{2} y y_{0}-b^{2} x_{0}^{2}+a^{2} y_{0}^{2}=0\)
\(\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y}{b^{2}}-\left(\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}\right)=0\)
\(\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y}{b^{2}}-1=0\)
\(\Rightarrow \frac{x x_{0}}{a^{2}}-\frac{y y}{b^{2}}=1\)
Then, slope of normal at \( \left(x_{0}, y_{0}\right) \)
\(=\frac{-1}{\text { Slope of the tangent at }\left(x_{0}, y_{0}\right)}=\frac{-a^{2} y_{0}}{b^{2} x_{0}}\)
Therefore, the equation of the normal at \( \left(x_{0}, y_{0}\right) \) is
\(y-y_{0}=\frac{-a^{2} y_{0}}{b^{2} x_{0}}\left(x-x_{0}\right)\)
\(\Rightarrow \frac{y-y_{0}}{a^{2} y_{0}}=\frac{-(x-x_0)}{b x_{0}}\)
\(\Rightarrow \frac{y-y_{0}}{a^{2} y_{0}}+\frac{x-x_{0}}{b x_{0}}=0\)
Therefore, the equation of the normal at \( \left(x_{0}, y_{0}\right) \) is \( \frac{y-y_{0}}{a^{2} y_{0}}+\frac{x-x_{0}}{b x_{0}}=0 \).

It is given that \( y=\sqrt{3 x-2} \)
Then, the equation of the tangent at any given 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 \( 4 x-2 y+5=0 \)
\(\Rightarrow y=2 x+\frac{5}{2}\)
\( \Rightarrow \) slope of the line \( =2 \)
Now, the tangent to the given curve is parallel to the line \( 4 x-2 y+5=0 \) if the slope of the tangent \( = \) the slope of the line
\(\frac{3}{2 \sqrt{3 x-2}}=2\)
\(\Rightarrow \sqrt{3 x-2}=\frac{3}{4}\)
\(\Rightarrow 3 x-2=\frac{9}{16}\)
\(\Rightarrow 3 x=\frac{9}{16}+2=\frac{41}{16}\)
\(\Rightarrow x=\frac{41}{48}\)
When \( x=\frac{41}{48} \),
\(y=\sqrt{3\left(\frac{41}{48}\right)-2}=\sqrt{\frac{41}{16}-2}=\sqrt{\frac{41-32}{16}}=\sqrt{\frac{9}{16}}=\frac{3}{4}\)
Then, Equation of the tangent passing through the point \( \left(\frac{41}{48}, \frac{3}{4}\right) \) is given by:
\(y-\frac{3}{4}=2\left(x-\frac{41}{48}\right)\)
\(\Rightarrow \frac{4 y-3}{4}=2\left(\frac{48 x-41}{48}\right)\)
\(\Rightarrow 4 y-3=\frac{48 x-41}{48}\)
\(\Rightarrow 24 y-18=48 x-41\)
\(\Rightarrow 48 x-24 y=23\)
Therefore, the equation of the required tangent is \( 48 x-24 y=23 \)

It I given that the slope of the normal to the curve \( y=2 x^{2}+3 \sin x \),
Then, the slope of the tangent at \( x=0 \) is
\(\left.\left.\frac{d y}{d x}\right]_{x=0}=4 x+3 \cos x\right]_{x=0}=0+3 \cos 0=3\)
Therefore, the slope of the normal to the curve at \( x=0 \) is
\(=\frac{-1}{\text { Slope of the tangent at } x=0}=\frac{-1}{3}\)
Therefore, the slope of the normal to the curve \( y=2 x^{2}+3 \sin x \) at \( x=0 \) is \( \frac{-1}{3} \).

It is given that tangent to the curve \( y^{2}=4 x \)
Then differentiating with respect to \(x \), we have,
\(2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y}\)
Then, the equation of the tangent at any given point ( \( x, y \) ) is given by, \( \frac{d y}{d x}=\frac{2}{y} \)
The given line is \( y=x+1 \)
\( \Rightarrow \) Slope of the line \( =1 \)
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.
Then, we have,
\(\frac{2}{y}=1\)
\(\Rightarrow y=2\)
Now, \( y=x+1 \)
\(\Rightarrow x=y-1\)
\(\Rightarrow x=2-1=1\)
Therefore, the line \( y=x+1 \) is a tangent to the given curve at the point \( (1 \), 2).