Ncert Solutions Maths class 12th

(i) we have, $y=x^4-6x^3+13x^2-10x+5$

slope of tangent, $\frac{dy}{dx}=4x^3-18x^2+26x-10$

${\frac{dy}{dx}|}_{(x,y)=(0,5)}=-10.$

slope of normal $=\frac{-1}{-10}=\frac{1}{10}$

Hence eqⁿ of tangent at (0, 5) is

$y-5=-10(x-0)\to 10x+y-5=0$

And eqⁿ of normal at (0, 5) is

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

$\Rightarrow 10y-50=x$

$\Rightarrow x-10y+50=0$

(ii) We have, $y=x^4-6x^3+13x^2-10x+5$

Slope of tangent, $\frac{dy}{dx}=4x^3-18x^2+26x-10.$

${\frac{dy}{dx}|}_{(x,y)=(1,3)}=4{(1)}^3-18{(1)}^2+26(1)-10$

$=4-18+26-10$

= 30 28

= 2

Slope of normal $=-\frac{1}{2}$

Hence eqⁿ of tangent at (1, 3) is

$y-3=2(x-1)$

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

$\to 2x-y+1=0$

And eqⁿ of normal at (1,3) is

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

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

$\to x+2y-7=0$

(iii) We have, $y=x^3$

Slope of tangent,&nb

Diffrentiating $\frac{x^2}{9}+\frac{y^2}{16}=1.$ wrt. X we get,

$\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$

$\Rightarrow \frac{2ydy}{16dx}=-\frac{2x}{9}$

$\Rightarrow \frac{dy}{dx}=\frac{-16x}{9y}$

(i) When the tangent is to x-axis, the slope of tangent is 0

ie, $\frac{dy}{dx}=0$

$\Rightarrow \frac{-16x}{9y}=0$

$\Rightarrow x=0$ putting this in the eqⁿ of curve. We get,

$\to \frac{0^2}{9}+\frac{y^2}{16}=1$

$\Rightarrow y^2=16$

$\Rightarrow y=\pm 4.$

The point at which the tangents are parallel to x-axis are $(0,4)and(0,-4)$

(ii) When the tangent is parallel to y-axis, the slope of the normal is 0.

ie, $\frac{-1}{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}=0$

$\Rightarrow -\frac{dx}{dy}=0$

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

$\to y=0$

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

Slope of tangent to the curve is $\frac{dy}{dx}=\frac{-1}{\left(x^2-2x+3\right)}=\frac{d}{dx}\left(x^2-2x+3\right)$

$=\frac{-(2x_1-2)}{{\left(x^2-2x+3\right)}^2}$

Given, $\frac{dy}{dx}=0$

$\to \frac{-(2x-2)}{{\left(x^2-2x+3\right)}^2}=0$

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

$\Rightarrow x=1$

When $x=1,y=\frac{1}{1^2-2\times 1+3}=\frac{1}{1-2+3}=\frac{1}{2}$

The point of contact of the tangent to the curve is $\left(1,\frac{1}{2}\right)$

The eqⁿ of the line is $y-\frac{1}{2}=0(x-1)$

$\to y=\frac{1}{2}$

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

Slope of tangent to the curve is $\frac{dy}{dx}=\frac{-1}{{(x-3)}^2}.$

Given,  $\frac{dy}{dx}=2$

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

$\Rightarrow {(x-3)}^2=-\frac{1}{2}$ which is not possible

we conclude that there is no possible tangent to the given curve with slope = 2.

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

Slope of tangent to the given curve is $\frac{dy}{dx}=\frac{-1}{{(x-1)}^2}$

Given that, slope of tangent = 1.

$\to \frac{-1}{{(x-1)}^2}=-1.$

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

$\Rightarrow x-1=\pm 1$

$\to x=1\pm 1.$

ie, X=1+1$$ or $x=1-1$

$\Rightarrow x=2$ or $x=0$

When $x=2,y=\frac{1}{2-1}=1$

and when $x=0,y=\frac{1}{0-1}=-1.$

Hence, the point of contact of the tangents are $(2,1)and(0,-1)$

The reqd. eqⁿ of line are $y-1=(-1)(x-2)\{\begin{array}{cc}\hfill ?y-y_0=m\left(x-x_0\right)\hfill & \hfill eqn{}^{} of line \hfill \end{array}$

and $y-(-1)=(-1)(x-0).$

$\Rightarrow y-1=-x+2$ and $y+1=-x$

$\to x+y-3=0$ and $x+y+1=0.$

The given eqⁿ of the curve is $y=x^3-11x+5$

slope of tangent to the curve $\frac{dy}{dx}=3x^2-11$

Then eqⁿ of tangent is $y=x-11$ $\Rightarrow x-y-11=0$ which gives us slope $=\frac{-1}{-1}=1$

So, $3x^2-11=1$

$\Rightarrow 3x^2=1+11=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

When x = 2, $y=2^3-11(2)+5=8-22+5=-9.$

And when x = 2, $y={(-2)}^3-11(-2)+5=-8+22+5=19.$

The point $(2,-9)$ when put into $y=x-11.$ we get

$-9=2-11$

$\Rightarrow -9=-9$ which is true.

and the point $\left(-2,19\right)$ when put into $y=x-11$ gives

Let the point joining the chord be $(2,0)(4,4)$

Then slope of the chord $=\frac{4-0}{4-2}\left\{?Slope=\frac{y_2-y_1}{x_2-x_1}\right\}$

$=\frac{4}{2}$

= 2

The given eqⁿ of the curve $y={(x-2)}^2$ 

slope of the tangent to the curve $\frac{dy}{dx}=2(x-2).$

Given that, the tangent is parallel to the chord PQ.

slope of tangent = slope of PQ.

$\Rightarrow 2(x-2)=2.$

$\Rightarrow x=1+2$

$\Rightarrow x=3.$

and $y={(3-2)}^2=1^2=1.$

The required point on curve is $(3,1)$

The given eqⁿ of the curve is $y=x^3-3x^2-9x+7.$

slope of tangent to the given curve, $\frac{dy}{dx}=3x^2-6x-9$

when the tangent is parallel to x-axis $\frac{dy}{dx}=0$

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2+x-3x-3=0$

$\to x(x+1)-3(x+1)=0$

$\to (x+1)(x-3)=0$

$$ x = 3 or x = -1

When x = 3, $y=3^3-3{(3)}^2-9(3)+7=27-27-27+7=-20$

And when x = -1 $y={(-1)}^3-3{(-1)}^2-9(-1)+7=-1-3+9+7=12$

Hence, the required points are $(3,-20)(-1,12)$

The given eqⁿ of the curves are

$x=1-asin\theta y=bco{s}^2\theta$

so,  $\frac{dx}{d\theta }=-acos\theta \frac{dy}{d\theta }=-2bcos\theta sin\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-2bcos\theta sin\theta }{acos\theta }=\frac{2b}{a}sin\theta$

Slope of tangent to curve at $\theta =\frac{\pi }{2}$ is ${\frac{dy}{dx}|}_{\theta =\frac{\pi }{2}}$

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

$=\frac{2b}{a}$

Hence, slope of normal to curve $=\frac{-1}{2b/a}=-\frac{a}{2b}$

The Equation of the given curve are

$x=aco{s}^3\theta y=asi{n}^3\theta$

So,  $\frac{dx}{d\theta }=3aco{s}^2\theta (-sin\theta )=-3aco{s}^2\theta sin\theta .$

and $\frac{dy}{d\theta }=3asi{n}^2\theta cos\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{3asi{n}^2\theta cos\theta }{-3aco{s}^2\theta sin\theta }=-tan\theta$

So,  ${\frac{dy}{dx}|}_{x=\pi /4}=-tan\frac{\pi }{4}=-1$ which is the slope of the tanget to the curve.

Now, required slope of normal to the curve =$-1$
$\frac{\mathrm{Slopeoftangent}}{}$ 
$\frac{}{}=\frac{-1}{-1}=1$