Question

The pole of the line $lx+my+n=0$ with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

• A. $(\frac{a^{2}l}{n},\frac{b^{2}m}{n})$
• B. $(-\frac{a^{2}l}{n},\frac{b^{2}m}{n})$
• C. $(\frac{a^{2}l}{n},-\frac{b^{2}m}{n})$
• D. $(-\frac{a^{2}l}{n},-\frac{b^{2}m}{n})$

Answer 1

The correct option is B $(-\frac{a^{2}l}{n},\frac{b^{2}m}{n})$

Let $P(x_1,y_1)$ be the pole of the line $lx+my+n=0$ with respect ot the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Then the equation of the polar is
$\frac{{xx}_1}{a^2}-\frac{{yy}_1}{b^2}=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$
Since $(x_1,y_1)$ is the pole of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(ii\right)$
Clearly $\left(i\right)$ and $\left(ii\right)$ represent the same line. Therefore,
$\therefore \frac{x_1}{a^{2}l}=\frac{{-y}_1}{b^{2}m}=\frac{1}{-n}$
$x_1=\frac{{-a}^{2}l}{n},y_1=\frac{b^{2}m}{n}$
Therefore, the pole of the given line with respect of the given hyperbola is
$(-\frac{a^{2}l}{n},\frac{b^{2}m}{n})$
Hence, option 'B' is correct.