Question

The coordinates of a point on the hyperbola $\frac{x^2}{24}-\frac{y^2}{18}=1$ which is nearest to the line $3x+2y+1=0$ are?

• A. $(6,3)$
• B. $(-6,-3)$
• C. $(6,-3)$
• D. $(-5,3)$

Answer 1

Answer: (C)

$(6,-3)$

The shortest distance would be perpendicular to the line

$\Rightarrow 3x+2y+1=0$

$\Rightarrow 2y=-3x-1$

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

A point on hyperbola that had same slope could be at minimum distance.

Slope of hyperbola by differentiating

$\Rightarrow \frac{x^2}{24}-\frac{y^2}{18}=1$

$\Rightarrow \frac{1}{24}\left(2xdx\right)-\left(\frac{1}{18}\right)\left(2ydy\right)=0$

$\Rightarrow \frac{xdx}{12}=\frac{ydy}{9}$

$\Rightarrow \frac{dx}{dy}=\frac{3x}{4y}=\frac{-3}{2}$

$\Rightarrow 6x=-12y$

$\therefore x=-2y$

Substituting in given hyperbola equation -

$\Rightarrow \frac{{\left(2y\right)}^2}{2y}-\frac{y^2}{18}=1$

$\Rightarrow \frac{y^2}{6}-\frac{y^2}{18}=1$

$\Rightarrow \Rightarrow \left(\frac{3}{18}\right).y^2-\left(\frac{9}{18}\right)y^2=1$

$\Rightarrow \left(\frac{2}{18}\right)y^2=1$

$\therefore y^2=9$

$y=+3$ and $y=-3$

Then, $\frac{x^2}{24}-\frac{9}{18}=1$

$\Rightarrow \frac{x^2}{24}=\frac{3}{2}$

$\therefore x^2=36$

$x=6$ and $x=-6$

so, point are $(-6,3)$ and $(6,-3)$

perpendicular line that would be shortest distance to original line $=\frac{2}{3}$

we have slope and point so the line ,

$\Rightarrow y-3=\frac{2}{3}[x-(-6)]$

$\Rightarrow y-3=\frac{2}{3}(x+6)$

$\Rightarrow y=\frac{2}{3}x+7.$

Hence, solved.