Question

The foot of the perpendicular from the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ on its asymptote is :

• A. $\left(\frac{4}{5},\frac{3}{5}\right)$
• B. $\left(\frac{3}{5},\frac{4}{5}\right)$
• C. $\left(\frac{16}{5},\frac{9}{5}\right)$
• D. $\left(\frac{16}{5},\frac{12}{5}\right)$

Answer 1

Answer: (D)

$\left(\frac{16}{5},\frac{12}{5}\right)$

Given hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$

So we see that for the hyperbola $a=4,b=3$

The equation of asymptote when the center of hyperbola is at origin is given by $y=\pm \frac{b}{a}x$

So in this case, equation of asymptote is $y=\pm \frac{3}{4}x$...........1

Focus of hyperbola $e=\sqrt{\frac{a^2+b^2}{a^2}}$

Putting values we get $e=\sqrt{\frac{25}{16}}=\frac{5}{4}$

So focus of hyperbola is $(\pm ae,0)=(\pm 5,0)$

The foot of perpendicular from the focus on asymptote should satisfy equation 1.

slope of asymptote $=\frac{3}{4}$ so slope of the perpendicular is $=\frac{-4}{3}$

so equation of perpendicular passing through the focus of hyperbola is $(y-0)=\frac{-4}{3}(x-5)\Rightarrow 3y+4x=20$.........2

So the required point is the point of intersection of 1 and 2.

Solving 1 and 2 we get $x=\frac{16}{5},y=\frac{12}{5}$