Question

Pole of a straight line $3x+5y+7=0$ with respect to the parabola \[{{y}^{2}}=12x\] is
(a) \[\left( \dfrac{5}{3},-8 \right)\]
(b) \[\left( \dfrac{5}{3},-10 \right)\]
(c) \[\left( -\dfrac{7}{3},10 \right)\]
(d) \[\left( \dfrac{7}{3},-10 \right)\]

Answer 1

Hint: The pole of a straight line \[lx+my+n=0\] to parabola \[{{y}^{2}}=4ax\] is given by \[\left( \dfrac{n}{l},-\dfrac{2am}{l} \right)\]. We can directly find the pole by substituting the known values.

Complete step-by-step solution -
The equation of the line is given as $3x+5y+7=0$ and the equation of the parabola is given as \[{{y}^{2}}=12x\].
Let us first look at the equation for the pole of a straight line of the general form \[lx+my+n=0\]with respect to the general form of parabola \[{{y}^{2}}=4ax\]. The pole is obtained as \[\left( \dfrac{n}{l},-\dfrac{2am}{l} \right)\].
Now the pole of the straight line $3x+5y+7=0$ to the parabola \[{{y}^{2}}=12x\] can be computed. On comparing the general equation of the line \[lx+my+n=0\], we get \[l=3,m=5,n=7\].
On comparing with the general equation of the parabola \[{{y}^{2}}=4ax\], we get
\[\begin{align}
  & 4a=12 \\
 & a=3 \\
\end{align}\]
Therefore on substituting \[l=3,m=5,n=7\] and \[a=3\], we get the pole as
$\begin{align}
  & \left( \dfrac{7}{3},-\dfrac{2\times 3\times 5}{3} \right) \\
 & \left( \dfrac{7}{3},-\dfrac{30}{3} \right) \\
 & \left( \dfrac{7}{3},-10 \right) \\
\end{align}$
Therefore, the pole is obtained as \[\left( \dfrac{7}{3},-10 \right)\].
Hence, the correct answer is option (d).

Note: A pole is defined as the point of intersection of the chords of contact to the tangents of a parabola. There is a chance that the student might try substituting the given options in the equations of the straight line as well as the parabola to check if it satisfies both the equations. The direct substitution will not lead to the answer, so it is advised to recollect the coordinates of the pole, i.e. \[\left( \dfrac{n}{l},-\dfrac{2am}{l} \right)\].