⇒Application Of Derivative

Correct Option is (C)

$3x^2-y^2=8$

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& 6x-2y\frac{dy}{dx}=0\\ \therefore & \frac{dy}{dx}=\frac{3x}{y}\end{array}$

$\therefore$ Slope of the tangent to the curve is $\frac{3x}{y}$.

$\therefore$ Slope of the normal is $\frac{-y}{3x}$.

It is parallel to line $x+3y=10\Rightarrow$ slope $=-\frac{1}{3}$

$\therefore \frac{-y}{3x}=\frac{-1}{3}\Rightarrow x=y$

$\therefore$ When $x=y$, equation of the curve becomes $3x^2-x^2=8$

$\therefore x^2=4$

$\therefore x=2,-2\Rightarrow y=2,-2$

$\therefore (2,2)$ and $(-2,-2)$ are the points of contact of the normal and the curve.

$\begin{array}{rl}\therefore & \text{ Equations are }(y-2)=\frac{-1}{3}(x-2)\text{ or }\\ & (y+2)=\frac{-1}{3}(x+2)\\ & \text{ i.e., }x+3y-8=0\text{ or }x+3y+8=0\end{array}$

Correct Option is (A)

$\begin{array}{rl}& y^2=6x\text{.... (i)}\\ & \Rightarrow 2y\frac{dy}{dx}=6\Rightarrow \frac{dy}{dx}=\frac{3}{y}\\ & \text{ Also, }9x^2+b{y}^2=16\\ & \Rightarrow 18x+2by\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-9x}{by}\end{array}$

As given curves intersect each other at righ angle, their tangents also intersect at right angles.

$\begin{array}{rl}& \frac{3}{y}\times \frac{-9x}{\mathrm{b}y}=-1\\ & \Rightarrow \mathrm{b}{y}^2=27x\end{array}$

$\begin{array}{rl}\therefore (i)& \Rightarrow b(6x)=27x\\ & \Rightarrow b=\frac{9}{2}\end{array}$

Correct Option is (A)

$y=2x^2$

$\therefore$ Slope of the tangent to this curve is

$\begin{array}{ll}& \frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{m}}_1=4x\\ \therefore & \text{ at }(1,1),{\mathrm{m}}_1=4\\ & x=2y^2\end{array}$

$\therefore$ Slope of the tangent to this curve is $\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{m}}_2=\frac{1}{4y}$

$\therefore \text{ at }(1,1),{\mathrm{m}}_2=\frac{1}{4}$

Let $\theta$ be the angle between two tangents.

$\begin{array}{rl}& \therefore \tan \theta =\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|=\left|\frac{4-\frac{1}{4}}{1+4\times \frac{1}{4}}\right|=\frac{15}{8}\\ & \therefore \theta ={\tan }^{-1}\left(\frac{15}{8}\right)\end{array}$

Correct Option is (D)

Given curve is $y=\sqrt{9-2x^2}$

If ordinate and abscissa are equal, we get $y=x$.

$\therefore$ Equation of the curve becomes $x^2=9-2x^2$

$\Rightarrow x=\pm \sqrt{3}$

If $x=-\sqrt{3}$, then $y=\sqrt{9-2(3)}=\sqrt{3}$

In this case, $x\ne y$.

Hence, $x\ne -\sqrt{3}$

$\therefore x=\sqrt{3}$ and $y=\sqrt{3}$

$\therefore$ Slope of the tangent to the given curve is $2y\frac{\mathrm{d}y}{\mathrm{d}x}=-4x$

$\therefore$ at $(\sqrt{3},\sqrt{3}),\frac{\mathrm{d}y}{\mathrm{d}x}=-2$

$\therefore$ Equation of the required tangent is

$\begin{array}{rl}& (y-\sqrt{3})=-2(x-\sqrt{3})\\ & \text{ i.e., }2x+y-3\sqrt{3}=0\end{array}$

Correct Option is (B)

$\begin{array}{ll}& xy=1\\ \therefore & y=\frac{1}{x}\\ \therefore & y^{\prime }=\frac{-1}{x^2}\\ \therefore & \text{ Slope of the normal }=x^2\\ & \text{ Slope of the line }\mathrm{a}x+\mathrm{b}y+\mathrm{c}=0\text{ is }\frac{-\mathrm{a}}{\mathrm{b}}.\end{array}$

Since the line $\mathrm{a}x+\mathrm{b}y+\mathrm{c}=0$ is a normal to the curve $xy=1$,

$x^2=-\frac{a}{b}$

For this condition to hold true, either $\mathrm{a}<0,\mathrm{b}>0$ or $\mathrm{b}<0,\mathrm{a}>0$