Question

For a standard hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Match the following.

$\begin{array}{cc}\text{Column 1}& \text{Column 2}\\ 1.a^2>b^2& \text{P.Director circle is real}\\ 2.a^2=b^2& \text{Q.Director circle is imaginary}\\ 3.a^2<b^2& \text{R.Centre is the only point from which}\\ & \text{two perpendicular tangents can be drawn on the}\\ & \text{hyperbola}\end{array}$

• A. $1-P,2-Q,3-R$
• B. $1-R,2-Q,3-P$
• C. $1-P,2-R,3-Q$
• D. $1-Q,2-P,3-R$

Answer 1

The correct option is C

$1-P,2-R,3-Q$

Locus of point of intersection of the perpendicular tangents of hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or director circle of hyperbola is
$x^2+y^2=a^2-b^2$
We can clearly see that if $a^2>b^2$, then radius of the circle will have some finite value.

$\Rightarrow$ the director circle is real.

when $a^2=b^2$, then radius of the director circle is zero and it reduces to point circle at origin. In this case

centre is the only point from which two perpendicular tangents can be drawn on the curve.

If $a^2<b^2$ the radius of the director circle is imaginary, so that there is no such circle and so no pair of

tangents at right angle can be drawn to the hyperbola.