Question

The locus of the foot of perpendicular from the focus on any tangent to the hyperbola $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is:

• A. $x^2+y^2=a^2+b^2$
• B. $x^2+y^2=a^2-b^2$
• C. $x=0$
• D. $x^2+y^2=a^2$

Answer 1

Answer: (A)

$x^2+y^2=a^2+b^2$

The line $y=mx\pm \sqrt{(a^2{m}^2+b^2)}$ is a tangent to the given ellipse for all $m$.

Suppose it passes through $(h,k)$

$\Rightarrow k-mh=\sqrt{(a^2{m}^2+b^2)}$

$\Rightarrow k^2+m^2{h}^2-2kmh=a^2{m}^2+b^2$

$\Rightarrow m^2(h^2-a^2)-2hkm+k^2-b^2=0.$

If the tangents are at right angles, then $m_1{m}_2=-1.$

$\Rightarrow (k^2-b^2)/(h^2-a^2)=-1$

$\Rightarrow h^2+k^2=a^2+b^2$.

Hence the locus of the point $(h,k)$ is $x^2+y^2=a^2+b^2$ which is a circle.