Question

The distance of the point on the curve $3x^2-4y^2=72$ nearest to the line $3x+2y+1=0$ is

• A. $\frac{\sqrt{13}}{11}$
• B. $\frac{9}{\sqrt{13}}$
• C. $\sqrt{13}$
• D. $\frac{11}{\sqrt{13}}$

Answer 1

The correct option is D $\frac{11}{\sqrt{13}}$

$3x^2-4y^2=72\to \left(1\right)$ $\&$ $3x+2y+1=0\to \left(2\right)$

Differentiation $\left(1\right)$ w.r.t $x$

$\Rightarrow 6x-8y.\frac{dy}{dx}=0$

${\left|\frac{dy}{dx}\right|}_{(x_1,y_1)}=\frac{3}{4}\frac{x_1}{y_1}$

slope of line \left(2\right) =\dfrac{-3}{2}$

slope -

$\Rightarrow \frac{3}{4}\frac{x_1}{y_1}=\frac{-3}{2}$

$x_1=-2y_1\to \left(3\right)$

putting the value in $\left(1\right)$

$\Rightarrow 3\left(4y_1^2\right)-4y_1^2=72$

$\Rightarrow 12y_1^2-4y_1^2=72$

$8y_1^2=72$

$y_1^2=9$

$\Rightarrow y_1=\pm 3$

when $y_1=3$ $x=-6$

$y_1=-3$ $x=6$

so the points are $(-6,3)\&(6,-3)$

at point $(-6,3)$

$\Rightarrow$ Distance $=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$ $3x+2y+1=0$ $a=3,b=2$

$=\frac{|3\times -6+2\times 3+1|}{\sqrt{3^2+2^2}}$

$=\frac{11}{\sqrt{13}}$

at point $(6,-3)$

$=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$=\frac{|18-6+1|}{\sqrt{13}}$

$=\frac{13}{\sqrt{13}}$

$=\sqrt{13}$

this value is discarded.

Hence, the value is $\frac{11}{\sqrt{13}}.$