KCET · Maths · Parabola
The locus of the mid point of the line joining the focus and any point on the parabola \(y^{2}=4 a x\) is a parabola with the equation of directrix as
- A \(\mathrm{x}+\mathrm{a}=0\)
- B \(2 x+a=0\)
- C \(\mathrm{x}=0\)
- D \(x=\frac{a}{2}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{x}=0\)
Step-by-step Solution
Detailed explanation
Let the coordinates of focus be \(S(a, 0)\).
Let any point on the parabola be \(P\) (at \({ }^{2}, 2\) at). Let the coordinates of mid point of \(P\) and \(S\) be \(\left(x_{1}, y_{1}\right)\).
\[
\begin{aligned}
& & \mathrm{x}_{1} &=\frac{\mathrm{a}+\mathrm{at}^{2}}{2}, \mathrm{y}_{1} \\
\Rightarrow & & \mathrm{at}^{2} &=2 \mathrm{x}_{1}-\mathrm{a}, \quad \mathrm{y}_{1} \\
\Rightarrow & \mathrm{a}\left(\frac{\mathrm{y}_{1}}{\mathrm{a}}\right)^{2} &=2 \mathrm{x}_{1}-\mathrm{a} \\
\Rightarrow & \mathrm{y}_{1}^{2} &=2 \mathrm{x}_{1} \mathrm{a}-\mathrm{a}^{2}
\end{aligned}
\]
Hence, the locus of the mid point is
\[
\mathrm{y}^{2}=2 \mathrm{a}\left(\mathrm{x}-\frac{\mathrm{a}}{2}\right)
\]
\(\therefore\) Equation of directrix is \(x-\frac{a}{2}=-\frac{a}{2}\)
\[
x=0
\]
Let any point on the parabola be \(P\) (at \({ }^{2}, 2\) at). Let the coordinates of mid point of \(P\) and \(S\) be \(\left(x_{1}, y_{1}\right)\).
\[
\begin{aligned}
& & \mathrm{x}_{1} &=\frac{\mathrm{a}+\mathrm{at}^{2}}{2}, \mathrm{y}_{1} \\
\Rightarrow & & \mathrm{at}^{2} &=2 \mathrm{x}_{1}-\mathrm{a}, \quad \mathrm{y}_{1} \\
\Rightarrow & \mathrm{a}\left(\frac{\mathrm{y}_{1}}{\mathrm{a}}\right)^{2} &=2 \mathrm{x}_{1}-\mathrm{a} \\
\Rightarrow & \mathrm{y}_{1}^{2} &=2 \mathrm{x}_{1} \mathrm{a}-\mathrm{a}^{2}
\end{aligned}
\]
Hence, the locus of the mid point is
\[
\mathrm{y}^{2}=2 \mathrm{a}\left(\mathrm{x}-\frac{\mathrm{a}}{2}\right)
\]
\(\therefore\) Equation of directrix is \(x-\frac{a}{2}=-\frac{a}{2}\)
\[
x=0
\]
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