Question

If the tangent and normal to $x^2-y^2=a^2$ at a point cut off intercepts $a_1,a_2$ and $b_1,b_2$ on the $x$-axis and $y$-axis respectively, then $a_1{a}_2+b_1{b}_2$ is

• A. $0$
• B. $1$
• C. $2$
• D. Dependent on position of the point

Answer 1

The correct option is A $0$

Let $P(x_1,y_1)$ is a point on $x^2-y^2=a^2$
Equation of the tangent at $P$ is $x{x}_1-y{y}_1=a^2\Rightarrow \frac{x}{\frac{a^2}{x_1}}+\frac{y}{\frac{-a^2}{y_1}}=1$
$\therefore a_1=\frac{a^2}{x_1},b_1=-\frac{a^2}{y_1}$

Equation of the normal at $P$ is $\frac{x}{x_1}+\frac{y}{y_1}=2\Rightarrow \frac{x}{2x_1}+\frac{y}{2y_1}=1$
$\therefore a_2=2x_1,b_2=2y_1$
$\Rightarrow a_1{a}_2+b_1{b}_2=\frac{a^2}{x_1}\cdot \left(2x_1\right)+(-\frac{a^2}{y_1})\left(2y_1\right)=0$