If the feet $A (a t_{1}^{2}, 2 a t_{1})$ and $B (a t_{2}^{2}, 2 a t_{2})$ are the ends of a focal chord of the parabola, then the locus of $P (h, k)$ is

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

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

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y^{2} = a (x - 3 a)

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y^{2} = 3 a (x - a)

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Solution

The correct option is C $y^{2} = a (x - 3 a)$
$A (a t_{1}^{2}, 2 a t_{1})$ and $B (a t_{2}^{2}, 2 a t_{2})$ are the ends of a focal chord then $t_{1} t_{2} = - 1$
From the cubic equation, we have,
$t_{1} t_{2} t_{3} = \frac{k}{a}$
$t_{3} = - \frac{k}{a}$
$t_{3}$ will satisfy the equation $a t^{3} + (2 a - h) t - k = 0$
$\Rightarrow - \frac{k^{3}}{a^{2}} - \frac{k}{a} (2 a - h) - k = 0$
$k^{2} = a (h - 3 a)$
Replacing $h$ by $x$ and $k$ by $y$ , we get the locus as
$y^{2} = a (x - 3 a)$
Hence, option C is correct

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Similar questions

y^{2} = 4 a x

y^{2} = 4 a x

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

A (a t_{1}^{2}, 2 a t_{1})

B (a t_{2}^{2}, 2 a t_{2})

y^{2} = 4 a x,

t_{1} t_{2} = - 1

The locus of the orthocenter of the triangle formed by the focal chord of the parabola $y^{2} = 4 a x$ and the normal drawn at its extermities is

y^{2} = 4 a x

y^{2} = 4 a x

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