Question

The point on the hyperbola $ 3{{x}^{2}}-4{{y}^{2}}=72 $ which is nearest to the line $ 3x+2y+1=0 $ is

• A. $ (-\text{ }6,3) $
• B. (6, 3)
• C. $ (-\text{ }6,-3) $
• D. $ (\text{ }6,-3) $
• E. $ (\sqrt{24},0) $

Answer 1

Answer: (A)

First we check which point satisfy the equation of hyperbola. All points in options are satisfied the equation of hyperbola $ 3{{x}^{2}}-4{{y}^{2}}=72 $ . Now, we find one-by-one the length of perpendicular from point on Ellipse to the line
$ 3x+2y+1=0 $ .
$ {{p}_{(-6,3)}}=\frac{11}{\sqrt{13}} $
$ {{p}_{(6,3)}}=\frac{25}{\sqrt{13}} $
$ {{p}_{(-6,-3)}}=\frac{23}{\sqrt{13}} $
$ {{p}_{(6,-3)}}=\frac{13}{\sqrt{13}} $ $ {{p}_{(\sqrt{24},0)}}=\frac{3\sqrt{24}+1}{\sqrt{13}} $
The minimum length is $ {{p}_{(-6,3)}} $ .
So, the point $ (-6,3) $ is nearest to the given line.