Question

The line $5x+12y=9$ touches the hyperbola $x^2-9y^2=9$ at the point.

• A. $(-5,\frac{4}{3})$
• B. $(5,\frac{-4}{3})$
• C. $(3,\frac{-1}{2})$
• D. None of these

Answer 1

The correct option is B $(5,\frac{-4}{3})$

The slope of the given line $5x+12y=9$ is $m=-\frac{5}{12}$
We know that if $m$ is the slope of a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then the coordinates of the point of contact are
$(\pm \frac{a^{2}m}{\sqrt{a^2{m}^2-b^2}},\pm \frac{b^2}{\sqrt{a^2{m}^2-b^2}})$
Given, equation of hyperbola is
$\frac{x^2}{9}-\frac{y^2}{1}=1$
Here, $a^2=9,b^2=1$
$\therefore$ The point of contact is
$(\pm \frac{9\times \frac{-5}{12}}{\sqrt{9\times \frac{25}{144}-1}},\pm \frac{1}{\sqrt{9\times \frac{25}{144}-1}})$
$=(mp5,\pm \frac{4}{3})=(-5,\frac{4}{3})$ and $(+5,-\frac{4}{3})$
But out of the two points only point $(5,-\frac{4}{3})$ lies on the line $5x+12y=9$.
Hence, required point is $(5,-\frac{4}{3})$.