Question

If the polars of $(x_1,y_1)$, $(x_2,y_2)$ with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are at right angles, then

• A. $\frac{x_1{x}_2}{y_1{y}_2}+\frac{a^2}{b^2}=0$
• B. $\frac{x_1{x}_2}{y_1{y}_2}=\frac{a^2}{b^2}$
• C. $\frac{x_1{x}_2}{y_1{y}_2}+\frac{a^4}{b^4}=0$
• D. $\frac{x_1{x}_2}{y_1{y}_2}=\frac{a^4}{b^4}$

Answer 1

The correct option is C $\frac{x_1{x}_2}{y_1{y}_2}+\frac{a^4}{b^4}=0$

The polar of$(x_1,y_1)$ and $(x_2,y_2)$ w.r t the hyperbola are given by,
$\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$ and $\frac{x{x}_2}{a^2}-\frac{y{y}_2}{b^2}=1$
Therefore slopes of these lines are $m_1=\frac{b^2}{a^2}.\frac{x_1}{y_1}$ and $m_2=\frac{b^2}{a^2}.\frac{x_2}{y_2}$
Given both lines are perpendicular
$\Rightarrow m_1\cdot m_2=-1\Rightarrow \frac{x_1{x}_2}{y_1{y}_2}=-\frac{a^4}{b^4}$
$\Rightarrow \frac{x_1{x}_2}{y_1{y}_2}+\frac{a^4}{b^4}=0$
Hence, option 'C' is correct.