Question

The points on the hyperbola $x^2$ - $y^2=a^2$ from where mutually perpendicular tangents can be drawn to the circle $x^2+y^2=a^2$ is/are

• A. $(a\sqrt{\frac{3}{2}},\frac{a}{\sqrt{2}})$
• B. $(-a\sqrt{\frac{3}{2}},\frac{a}{\sqrt{2}})$
• C. $(-a\sqrt{\frac{3}{2}},-\frac{a}{\sqrt{2}})$
• D. $(a\sqrt{\frac{3}{2}},-\frac{a}{\sqrt{2}})$

Answer 1

The correct option is A $(a\sqrt{\frac{3}{2}},\frac{a}{\sqrt{2}})$

The equation of the tangent to the circle is:

$y=mx+a\sqrt{1+m^2}$

Let $P(h,k)$ lies on the tangent,

$k=mh+a\sqrt{1+m^2}$

$k-mh=a\sqrt{1+m^2}$

Squaring both sides we get:

$k^2+{\left(mh\right)}^2-2kmh=a^2(1+m^2)$

$m^2(h^2-a^2)-2mhk+k^2-a^2=0$

This is a quadratic equation in $m$

Let roots be $m_1$ and $m_2$:

$m_1{m}_2=-1$ ..... ($\perp$ tangents)

$\frac{k^2-a^2}{(h^2-a^2)}=-1$

$h^2+k^2=2a^2$

So, the points will lie on:

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

Now the point of intersection with the hyperbola is :

$x^2-y^2=a^2$ ........ $\left(1\right)$

$x^2+y^2=2a^2$ ........ $\left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$ we get:

$x=a\sqrt{\frac{3}{2}}$

Similarly, $y=\frac{a}{\sqrt{2}}$