Ncert Solutions Maths class 12th

Given,  $y=2x^2+3sinx$

Slope of tangent,  $\frac{dy}{dx}=4x+3cosx$

${\frac{dy}{dx}|}_{x=0}=4*0+3*cos0=3$

So, slope of normal $=\frac{-1}{3}$

Option (D) is correct.

The given eqⁿ of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ______(1)

Differentiating eqn (1) wrt 'x' we get,

$\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$

$\Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{2x}{a^2}$

$\Rightarrow \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$

${\frac{dy}{dx}|}_{\left(x,y\right)}=\left(x_0,y_0\right)\frac{b^2{x}_0}{a^2{y}_0}$ is the reqd slope of tangent to the curve

So, eqⁿ of tangent at point $\left(x_0,y_0\right)$ is

$y-y_0=\frac{b^2{x}_0}{a^2{y}_0}\left(x-x_0\right).$

$\Rightarrow \frac{y{y}_0}{b^2}-\frac{{y_0}^2}{b^2}=\frac{x{x}_0}{a^2}-\frac{{x_0}^2}{a^2}$

$\Rightarrow \frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=\frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}$

As $\left(x_0,y_0\right)$ lies on the parabola given by eqn (1) we write,

$\frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}=1$

Hence, $\frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1$

The given eqⁿ of the curves are

$x=y^2$ ________(1)

and $xy=x$ ___________(2)

Differentiating eqn (1) and (2) wrt 'x' we get,

$$ $\frac{d{y}^2}{dx}=\frac{dx}{dx}$

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

$\to \frac{dy}{dx}=\frac{1}{2y}$ _________(3)

and $\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(x\right)$

$\to x\frac{dy}{dx}+y=0$

$\to \frac{dy}{dx}=-\frac{y}{x}$ _________(4)

Since the two curves cut each other at right angles we get,

$\left(\frac{1}{2y}\right)*\left(\frac{-y}{x}\right)=-1.$

$\to 2x=1$ ________(5)

The point of intersection can be solve from eqn (1) and (2), 

$y^2.y=k$

$\to y^3=k$

$\to y=k^{1/3}$

$\therefore x=y^2=r^{2/3}.$

Hence, using eqⁿ (5) we get,

$2{\left(r\right)}^{2/3}=1.$

$\Rightarrow {\left[2r^{2/3}\right]}^3=1^3$

$\Rightarrow 8r^2=1$

Hence proved

The given of the parabola is $y^2=4ax$

slope of tangent is given by

$\frac{d}{dx}{y}^2=\frac{d}{dx}4ax$

$\to 2y\frac{dy}{dx}=4a$

$\Rightarrow \frac{dy}{dx}=\frac{2a}{y}.$

${\frac{dy}{dx}|}_{\left(a{t}^2,2at\right)}=\frac{2a}{2at}=\frac{1}{t}$

so, slope of normal

so, slope of normal $=\frac{-1}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$t$}\right.\right)}=-t$

Hence eqⁿ of tangent at point $\left(a{t}^2,2at\right)$ is

$y-2at=\frac{1}{t}\left(x-a{t}^2\right)$

$\to ty-2a{t}^2=x-a{t}^2$

$\to x-ty-a{t}^2+2a{t}^2=0$

$x-ty+a{t}^2=0$

And eqⁿ of normal at point $\left(a{t}^2,2at\right)$ is

$\left(y-2at\right)=\left(-t\right)\left(x-a{t}^2\right)$

$\to y-2at=-tx+a{t}^3$

$\Rightarrow tx+y-2at+a{t}^3=0$

$\Rightarrow tx+y-at\left(2+t^2\right)=0$

The given eqⁿ of the curve is $y=x^3+2x+6$ _____(1)

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

so, slope of normal to the curve $=\frac{-1}{3x^2+2}$

Now, the line $x+14y+4=0\to y=-\frac{1}{14}x-\frac{4}{14}$ compared to $y=mx+c$ gives

slope of line = $-\frac{1}{14}$

As the normal is parallel to the line

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

$\to 3x^2+2=14$

$\Rightarrow 3x^2=12$

$\to x^2=4$

$\Rightarrow x=\pm 2$

When $x=2,y={(2)}^3+2(2)+6=8+4+6=18$

and when $x=-2,y={(-2)}^3+2(-2)+6=-8-4+6=-6$

The point of contact of the normal are (2, 18) and (-2, -6)

Hence the eqⁿ of normal are

$y-18=-\frac{1}{14}(x-2)$ and $y-(-6)=\frac{-1}{14}[x-(-2)]$

$\Rightarrow 14y-252=-x+2$ $\Rightarrow y+6=\frac{-1}{14}(x+2)$

$\Rightarrow x+14y-254=0$ $\Rightarrow 14y+84=-x-2$

$\Rightarrow x+14y+86=0.$

The given eqⁿ of curve is $a{y}^2=x^3$ _____(1)

Differentiating eqⁿ (1) wrt.x we get,

$2ay\frac{dy}{dx}=3x^2.$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ay}$ , slope of tangent

${\frac{dy}{dx}|}_{(x,y)=\left(a{m}^2,a{m}^3\right)}=\frac{3{\left[a{m}^2\right]}^2}{2a\left[a{m}^3\right]}=\frac{3a^2{m}^4}{2a^2{m}^3}=\frac{3m}{2}$

corresponding slope of normal $=\frac{-1}{(3m/2)}=-\frac{2}{3m}$

Hence, eqⁿ of normal at $\left(a{m}^2,a{m}^3\right)$ is

$y-a{m}^3=-\frac{2}{3m}\left(x-a{m}^2\right).$

$\Rightarrow 3my-3a{m}^4=-2x+2a{m}^2$

$\Rightarrow 2x+3my-3a{m}^4-2a{m}^2=0$

$\Rightarrow 2x+3my-a{m}^2\left(2+3m^2\right)=0$

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 $(0,0)and(3,27)$