MHT CET · Maths · Parabola
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 \(\left(\frac{-n}{l}, \frac{2 a m}{l}\right)\)
- C \(\left(\frac{n}{l}, \frac{-2 a m}{l}\right)\)
- D \(\left(\frac{n}{l}, \frac{2 a m}{l}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{n}{l}, \frac{-2 a m}{l}\right)\)
Step-by-step Solution
Detailed explanation
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 \(\left\{\frac{n}{l}, \frac{-2 a m}{l}\right\} .\)
\(
y y_{1}=2 a\left(x+x_{1}\right)
\)
\(
\Rightarrow \quad 2 a x_{1}-y y_{1}+2 \alpha x=0 \ldots(\mathrm{i})
\)
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 \(\left\{\frac{n}{l}, \frac{-2 a m}{l}\right\} .\)
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