Through the vertex O of a parabola ^ 2 =4 x , chords OP and O Q are...

Through the vertex

\mathrm{O}

of a parabola

\mathrmy^{2}=4 \mathrm{x}

, chords OP and

O Q

are drawn at right angles to one another. The locus of the middle point of

\mathrm{PQ}

Options

A $y^{2}=2 x+8$
B $\quad y^{2}=x+8$
C $y^{2}=2 x-8$
D $\quad y^{2}=x-8$

Correct answer

C. $y^{2}=2 x-8$

Step-by-step solution

Given parabola is

\mathrm{y}^{2}=4 ...(1) \mathrm{x}

Let

\mathrm{P} \equiv\left(\mathrm{t}_{1}^{2}, 2 \mathrm{t}_{1}\right)

and

\mathrm{Q} \equiv\left(\mathrm{t}_{2}^{2}, 2 \mathrm{t}_{2}\right)

Slope of

\mathrm{OP}=\frac{2 \mathrm{t}_{1}}{\mathrm{t}_{1}^{2}}=\frac{2}{\mathrm{t}_{1}}

and slope of

\mathrm{OQ}=\frac{2}{\mathrm{t}_{2}}

Since

\mathrm{OP} \perp \mathrm{OQ}, \therefore \frac{4}{\mathrm{t}_{1} \mathrm{t}_{2}}=-1

\mathrm{t}_{1} \mathrm{t}_{2}=-4 \ldots(2)

Let

\mathrm{R}(\mathrm{h}, \mathrm{k})

be the middle point of

\mathrm{PQ},

then

\mathrm{h}=\frac{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}}{2} \quad \ldots(3) \quad

and

\quad

\mathrm{k}=\mathrm{t}

_{1}+\mathrm{t}_{2} \ldots(4)

From

(4), \mathrm{k}^{2}=\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}=2 \mathrm{~h}-8 \quad[\mathrm{From}

(2) and (3)] Hence locus of

\mathrm{R}(\mathrm{h}, \mathrm{k})

\mathrm{y}^{2}=2 \mathrm{x}-8

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