Question

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

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

Answer 1

The correct option is D

(-6,3)

It is that point on the hyperbola where the tangent is parallel to the given line $3x+2y+1=0$, whose slope is $\frac{-3}{2}$.

We know that for a given slope $m$, there are two tangents to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the contact points of these tangents and the hyperbola are

$P(\frac{-a^{2}m}{\pm \sqrt{a^2{m}^2-b^2}},\frac{-b^2}{\pm \sqrt{a^2{m}^2-b^2}})$

Putting $a^2=24,b^2=18$, and $m=\frac{3}{2}$, we have
$P(-6,3)$ and $p(6,-3)$
Now, we can see that $(-6,3)$ in nearer to the line $3x+2y+1=0$