Question

Find the point of the hyperbola $\frac{x^2}{24}-\frac{y^2}{18}=1$ which is nearest to the line $3x+2y+1=0$

• A. (6,-3)
• B. (6,6)
• C. (3,6)
• D. None of these

Answer 1

Answer: (A)

(6,-3)

Given equation of hyperbola is $\frac{x^2}{24}-\frac{y^2}{18}=1$

Here, $a=2\sqrt{6},b=3\sqrt{2}$

Let $P(2\sqrt{6}\sec \theta ,3\sqrt{2}\tan \theta )$ be any point on the hyperbola .

Let p be its distance from the line $3x+2y+1=0.$

$p=\left|\frac{6\sqrt{6}\sec \theta +6\sqrt{2}\tan \theta +1}{\sqrt{13}}\right|$

$\Rightarrow p^2=\frac{1}{13}(6\sqrt{6}\sec \theta +6\sqrt{2}\tan \theta +1{)}^2$

Let $z=p^2$

Now p will be max. or min. if Z is max. or min.

$\therefore z=\frac{1}{13}(6\sqrt{6}\sec \theta +6\sqrt{2}\tan \theta +1{)}^2$

$\frac{dz}{d\theta }=2p\frac{dp}{d\theta }=0$

$\Rightarrow \frac{dp}{d\theta }=0$ ...as $p\ne 0$

$6\sqrt{6}\sec \theta \tan \theta +6\sqrt{2}{\sec }^2\theta =0$

$\Rightarrow 6\sqrt{2}[\sqrt{3}\sec \theta \tan \theta +{\sec }^2\theta ]=0$

$\Rightarrow \sqrt{3}\sin \theta +1=0$

$\Rightarrow \sin \theta =-\frac{1}{\sqrt{3}}$

$\therefore \cos \theta =\pm \sqrt{\frac{2}{3}}$

Hence, the points on the hyperbola are $(6,-3),(-6,3)$.

Required point is $(6,-3)$

Distance $=\sqrt{13}$ units