If from the vertex of a parabola a pair of chords be drawn at right angles to one another, & with these chords as adjacent sides a rectangle be constructed , then find the locus of the outer corner of the rectangle.

y^{2} = 4 a (x - 8 a)

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y^{2} = 2 a (x + 4 a)

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y^{2} = 2 a (x - 4 a)

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y^{2} = 4 a (x + 8 a)

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Solution

The correct option is A $y^{2} = 4 a (x - 8 a)$
Let parabola $y^{2} = 4 a x$
Let $P (a t_{1}^{2}, 2 a t_{1})$ , & $Q (a t_{2}^{2}, 2 a t_{2})$
OP & OQ are perpendicular
$\Rightarrow \frac{2}{t_{1}} \cdot \frac{2}{t_{2}} = - 1$
$t_{1} t_{2} = - 4$
Now diagonals of a rectangle bisect each other
$\frac{h}{2} = \frac{a t_{1}^{2} + a t_{2}^{2}}{2} \Rightarrow h = a (t_{1}^{2} + t_{2}^{2})$ .... (1)
$\frac{k}{2} = \frac{2 a t_{1} + a t_{2}}{2} \Rightarrow k = 2 a (t_{1} + t_{2})$ ....(2)
$\frac{k^{2}}{4 a^{2}} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}$
$\frac{k^{2}}{4 a^{2}} = \frac{h}{a} - 8$
Required locus is $y^{2} = 4 a (x - 8 a)$

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Similar questions

y^{2} = 4 a (x - 8 a) .

y^{2} = 4 a x

y^{2} = 4 a x

y^{2} = 4 a x

y^{2} = 2 a (x - 4 a)

P Q

t_{1}

t_{2}

t_{1} t_{2} = - 2

y^{2} = 4 a x

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