Question

Find the equation to the director circle of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

• A. x² + y² = a² - b²
• B. x² - y² = a² - b²
• C. x² + y² = a² + b²
• D. x² - y² = a² + b²

Answer 1

The correct option is A

x² + y² = a² - b²

Director circle is the locus of the point of intersection of the tangents which are right angled.

Let P(h,k) be the point of intersection of two perpendicular tangents.

Equation of pair of tangents is $s{s}_1=T^2$

$\Rightarrow (\frac{x^2}{a^2}-\frac{y^2}{b^2}-1)(\frac{h^2}{a^2}-\frac{k^2}{-b^2}-1)={(\frac{hx}{a^2}-\frac{ky}{b^2})}^2$

$\frac{x^2}{a^2}(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1)-\frac{y^2}{b^2}(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1)-(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1)$

$=\frac{h^2{x}^2}{a^4}+\frac{k^{2}y}{b^4}+1\frac{-2hkxy}{a^2{b}^2}+\frac{2ky}{b^2}+\frac{2hx}{a^2}$

$\frac{x^2}{a^2}(\frac{-k^2}{b^2}-1)-\frac{y^2}{b^2}(\frac{h^2}{a^2}-1)-(\frac{h^2}{a^2}-\frac{k^2}{b^2})+\frac{-2hkxy}{a^2{b}^2}-\frac{2ky}{b^2}+\frac{2hx}{a^2}=0$ ........(1)

Comparing this equation with standard pair of line eqution

$a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0$

Since, equation (1) represents two perpendicular line

$a+b=0$ or

Coefficient of $x^2+$ coefficient of $y^2=0$

$\frac{1}{a^2}(\frac{-k^2}{b^2}-1)-\frac{1}{b^2}(\frac{h^2}{a^2}-1)=0$

$\frac{-k^2}{a^2{b}^2}-\frac{1}{a^2}-\frac{h^2}{a^2{b}^2}+\frac{1}{b^2}=0$