Class 12th

The given eqⁿof the curve is $x^2+y^2-2x-3=0$ ________ (1)

Differentiating the given curve wrt.x we get,

$2x+2y\frac{dy}{dx}-2=0.$

$\Rightarrow 2y\frac{dy}{dx}=2-2x$

$\to \frac{dy}{dx}=\frac{1-x}{y}$ slope of tangent

Given, tangent is | to x-axis

ie,  $\frac{dy}{dx}=0$

$\Rightarrow \frac{1-x}{y}=0$

$\Rightarrow x=1$

Putting x = 1 in eqⁿ (1) we get,

$1^2+y^2-2\left(1_i\right)-3=0$

$\Rightarrow y^2-4=0$

$\Rightarrow y^2=4$

$\Rightarrow y=\pm 2$

Hence, the required points are (1,2) and (1, 2).

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

Slope of tangent, $\frac{dy}{dx}=12x^2-10x^4.$ ________(1)

Let P(x, y) be the required point at the tangent passing through the origin (0,0)

Then, $\frac{dy}{dx}=\frac{y-0}{x-0}=\frac{y}{x}$ _________(2)

So, from (1) and (2) we get,

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

$\Rightarrow y=12x^3-10x^5.$

Putting this value of y in the eqⁿ of curve we get,

$12x^3-10x^5=4x^3-2x^5.$

$\to 8x^3-8x^5=0$

$\Rightarrow 8x^3\left(1-x^2\right)=0$

$\Rightarrow x^3=0$ or $1-x^2=0$

$\Rightarrow x=0$ or $x=\pm 1.$

When, $x=0,y=4{(0)}^3-2{(0)}^5=0$

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

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

The required points are (0,0), (1,2) and (1,2)

The given eqⁿ of the curve is $y=x^3$ .

Slope of tangent,  $\frac{dy}{dx}=3x^2$

As, slope of tangent = y – coordinate of the point.

$\frac{dy}{dx}=y$

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

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

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

$\Rightarrow x=0x=3$

When $x=0,y=0$

and when $x=3·,y=3^3=27.$

The required points are $(0,0)and(3,27)$

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

Slope of tangent $\frac{dy}{dx}=21x^2$

${\frac{dy}{dx}|}_{x=2}=21{(2)}^2=21*4=84.$

and ${\frac{dy}{dx}|}_{x=-2}=21{(-2)}^2=21*4=84$

The tangent to the given curve at x = 2 and x = -2 are parallel.

The eqⁿof the given curve is $y=x^2-2x+7$

Slope of tangent, $\frac{dy}{dx}=2x-2$

(a) The line $2x-y+9=0\Rightarrow y=2x+9$ compared to $y=mx+c$ gives,

Slope of line = 2.

If the tangent of the curve is parallel to the line

$\frac{dy}{dx}=slope\; of\; line$

$\to 2x-2=2$

$\to x=\frac{4}{2}\Rightarrow x=2$

When $x=2,y={(2)}^2-2(2)+7=7$

Hence, the point of contact of the tangent is (2, 7)

The eqⁿ of tangent is $y-7=2(x-2)$

$\Rightarrow y-7=2x-4$

$\to 2x-y+3=0$

(b) The line $5y-15x=13\to y=\frac{15}{5}x+\frac{13}{5}\Rightarrow y=3x+\frac{13}{5}$

compared to $y=mx+c$ gives

slope of line = 3

As the tangent to the curve is ⊥ to the line.

$\frac{dy}{dx}=\; -1/slope\; opf\; line$

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

$\Rightarrow 6x-6=-1$

$\to 6x=5$

$\to x=\frac{5}{6}$

When $x=\frac{5}{6}$ we get $y={\left(\frac{5}{6}\right)}^2-2*\frac{5}{6}+7$

$=\frac{25}{36}-\frac{5}{3}+7$

$=\frac{25-60+252}{36}=\frac{217}{36}$

Hence, the point of contact of the tangent is $\left(\frac{5}{6},\frac{217}{36}\right)$

And eqⁿ of the tangent is

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

$\Rightarrow 3y-\frac{217}{12}=-x+\frac{5}{6}$

$\Rightarrow x+3y-\frac{217}{12}-\frac{5}{6}=0$

$\Rightarrow x+3y-\frac{217-10}{12}=0$

$\Rightarrow x+3y-\frac{227}{12}=0$

$\Rightarrow 12x+36y-227=0.$

(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, $\frac{dy}{dx}=3x^2$

${\frac{dy}{dx}|}_{(1,1)}=3{(1)}^2=3.$

And slope of normal $=-\frac{1}{3}$

Hence, eqn of tangent at (1, 1) is

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

$\Rightarrow y-1=3x-3$

$\to 3x-y-2=0$

And eqⁿ of normal at (1,1) is

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

$\to 3y-3=-x+1$

$\Rightarrow x+3y-4=0.$

(iv) We have, $y=x^2$

Slope of tangent $\frac{dy}{dx}=2x$

${\frac{dy}{dx}|}_{(0,0)}=0.$

So, eqⁿ of the tangent at (0,0) is

$(y-0)=0(x-0)$

$\Rightarrow y=0.$

ie, x- axis

Hence, the eqⁿ of normal is x = 0 ie, y-axis

(v) We have, $x=costy=sint$

$\frac{dx}{dt}=-sint·\frac{dy}{dt}=cost$

So, slope of tangent $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{cost}{-sint}=-cott$

${\frac{dy}{dx}|}_{t=\frac{\pi }{4}}=-cot\frac{\pi }{4}=-1.$

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

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$}\right.}=0$

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

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

$\to y=0$ , putting this in the eqⁿ of curve we get,

$\frac{x^2}{9}+\frac{y^2}{16}=1.$

$\to x^2=9$

$\to x=\pm 3$

The point at which the tangents are parallel to y-axis are $(3,0)and(-3,0)$

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

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.