Question

The pole of the line lx + my + n = 0 w.r.t the parabola \[{y^2} = {\text{ }}4ax\]will be :
A) $\left( {\dfrac{{ - n}}{l},\dfrac{{ - 2am}}{l}} \right)$
B) $\left( {\dfrac{{ - n}}{l},\dfrac{{2am}}{l}} \right)$
C) $\left( {\dfrac{n}{l},\dfrac{{ - 2am}}{l}} \right)$
D) $\left( {\dfrac{n}{l},\dfrac{{2am}}{l}} \right)$

Answer 1

Hint: Basically the polar of the parabola is the line which is same as of its tangent. And the pole of the line is the value of x and y in terms of its coefficients and constant terms.

Complete step by step solution: Here we first find the tangent of the parabola :
\[{y^2} = 4ax:\]
Which will be equals to :
\[y{y_1} = {\text{ }}2ax\left( {x + {x_1}} \right)\]
as \[{y^2}\]→ replaced by \[y{y_1}\]
and x → replaced by $\left( {\dfrac{{x + {x_1}}}{2}} \right)$
By replacing \[{y^2}\] and x by \[y{y_1}\] and $\left( {\dfrac{{x + {x_1}}}{2}} \right)$ respectively
\[y{y_1} = {\text{ }}2a\left( {x{\text{ }} + {\text{ }}{x_1}} \right)\].
So, it will be the tangent of the parabola.
The Equation of its tangents are :
\[y{y_1} = {\text{ }}2a\left( {x{\text{ }} + {\text{ }}{x_1}} \right)\]
\[\left( {2ax + 2a{x_1}-y{y_1}} \right) = 0\]–––––––– (i)
Which is same as the equation of parabolas polar line.
i.e., lx + my + n = 0 ––––––––– (ii)
Now comparing the co-efficient both lines, we get,
$\dfrac{{2a}}{l} = \dfrac{{ - y}}{m} = \dfrac{{2ax}}{n}$
Now, we will take first two parts :
$\left( {y = \dfrac{{ - 2am}}{l}} \right)$
As, $\left( {\dfrac{{2a}}{l} = \dfrac{{ - y}}{m}} \right)$ from here we $\left( {y = \dfrac{{ - 2am}}{l}} \right)$
$\left( {\dfrac{{2ax}}{n} = \dfrac{{ - y}}{m} = \dfrac{{2a}}{l}} \right)$
$x = \left( {\dfrac{{2an}}{{2al}}} \right)$
$\left( {x = \dfrac{n}{l}} \right)$
From here we get $\left( {x = \left( {\dfrac{n}{l}} \right)} \right)$
So, the pole of the line lx + my + c = 0 is the value of x and y i.e., option (C) is correct.
$\left( {x = \dfrac{n}{l}} \right),\& \left( {y = \dfrac{{ - 2am}}{l}} \right)$

Note: We will first find the equation of the parabola’s tangent by simply replacing \[{y^2}\] by \[y{y_1}\] and x by $\dfrac{{\left( {x + {x_1}} \right)}}{2}$ in the equation of \[{y^2} = {\text{ }}4ax\] and then compare the equation with the polar of the parabola to find its pole w.r.t parabola of any line in the terms of x and y.