A tangent is drawn at the point $P (α, β)$ on the parabola $y^{2} = 4 a x$ . This tangent meets a second parabola $y^{2} = 4 a (x + k)$ at A and B, what is the midpoint of AB?

$(α + k, β)$

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$(α - k, β - k)$

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$[\frac{α}{k}, \frac{β}{k}]$

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$(α, β)$

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Solution

The correct option is D
$(α, β)$

Lets first see how the given situation looks like

Here, for the tangent to intersect parabola $y^{2} = 4 a (x + k)$ note that k should be positive. This is because the second parabola should be outside the first one, i.e, the vertex of the second parabola should lie on the left of the first parabola.

The tangents equation can be given by,

$t y = x + a t^{2}$ - - - - - - (1)

where $(α, β) = (a t^{2}, 2 a t)$

Solving (1) and parabola $y^{2} = 4 a (x + k)$ we gets,

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

$= 4 a (y t - a t^{2} + k)$

i.e., $y^{2} + y (- 4 a t) + 4 a^{2} t^{2} - 4 a b = 0$

sum of roots = 4at

if $A \equiv (x_{1}, y_{1}) B (x_{2}, y_{2})$

$y_{1} + y_{2} = 4 a t$ - - - - - - (2)

$M i d p o i n t o f A B \equiv (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

$∴ y_{M P} = \frac{4 a t}{2} = 2 a t (u s i n g (2))$

$= β$

Also,

${[\frac{x t + a t^{2}}{t}]}^{2} = 4 a (x + k)$

$x^{2} + a^{2} t^{4} + 2 x a t^{2} = 4 a t^{2} x + k t^{2}$

$x^{2} + x (2 a t^{2} - 4 a t^{2}) + a 2 t^{4} - k t^{2} = 0$

$x^{2} + x (- 2 a t^{2}) + a^{2} t^{4} - k t^{2} = 0$

sum of the roots= $x_{1} + x_{2} = 2 a t^{2}$

$∴ \frac{x_{1} + x_{2}}{2} = a t^{2} = α$

$∴ m i d p o i n t o f A B = (α, β)$

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