Q77 / 150 · MHT CET 2012 · Mathematics
The pole of the line $L x+m y+n=0$ w.r.t. the parabola $y^{2}=4 a x$ will be
- A$\left(\frac{-n}{l}, \frac{-2 a m}{l}\right)$
- B$\left(\frac{-n}{l}, \frac{2 a m}{l}\right)$
- Correct answerC$\left(\frac{n}{l}, \frac{-2 a m}{l}\right)$
- D$\left(\frac{n}{l}, \frac{2 a m}{l}\right)$
Solution
Tangent of the parabola $y^{2}=4 a x$ at point $P\left(x_{1}, y_{1}\right)$ is
$$
y y_{1}=2 a\left(x+x_{1}\right)
$$
$$
\Rightarrow \quad 2 a x_{1}-y y_{1}+2 \alpha x=0 \ldots(\mathrm{i})
$$
which is also equation of the polar of the parabola $y^{2}=4 a x$. Same as the line
$$
b x+m y+n=0...(ii)
$$
On comparing both lines, we get
$$
\frac{2 a}{l}=\frac{-y}{m}=\frac{2 a x}{n}
$$
Taking first two parts
$$
\left(y=-\frac{2 a m}{l}\right)
$$
Taking first and last parts,
$$
\left(x=\frac{n}{l}\right)
$$
$\therefore$ Required pole of the line is $\left\{\frac{n}{l}, \frac{-2 a m}{l}\right\} .$
Hence, the correct answer is option: 'C'.