If one end of a focal chord of the parabola $y^{2} = 4 x$ is $(1, 2)$ , the other end lies on

x^{2} y + 2 = 0

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x y + 2 = 0

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x y - 2 = 0

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x^{2} + x y - y - 1 = 0

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Solution

The correct options are
A $x^{2} y + 2 = 0$
B $x y + 2 = 0$
D $x^{2} + x y - y - 1 = 0$
Given parabola is $y^{2} = 4 a x$ (where $a = 1)$
We know coordinates of end point of any focal chords are $P (a t^{2}, 2 a t)$ and $Q (\frac{a}{t^{2}}, - \frac{2 a}{t})$
Now let $P = (1, 2)$ so corresponding $t = 1$
So other end of the focal chord is $Q (1, - 2)$
Clearly point $Q$ lies on the curves in options $^{'} A^{'},^{'} B^{'},^{'} D^{'} .$

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