The tangent at the point $P (x_{1}, y_{1})$ to the parabola $y^{2} = 4 a x$ meets the parabola $y^{2} = 4 a (x + b)$ at $Q$ and $R$ , then the midpoint of $Q R$ is

(x_{1}, y_{1})

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(x_{1} + b, y_{1})

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(x_{1} + b, y_{1} + b)

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(x_{1} - b, y_{1} - b)

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Solution

The correct option is A $(x_{1}, y_{1})$
$y t = x + a t^{2}$
$h = (\frac{x_{2} + x_{3}}{2}), k = (\frac{y_{2} + y_{3}}{2})$
$y^{2} = 4 a (y t - a t^{2} + b)$
$y_{2} + y_{3} = 4 a t$
$\frac{(x + a t^{2})^{2})}{t^{2}} = 4 a x + 4 a b$
$x^{2} + 2 a t^{2} x + a^{2} t^{4} - 4 a b t^{2} - 4 a x t^{2} = 0$
$x_{3} + x_{2} = 2 a t^{2}$
$h = \frac{x_{3} + x_{2}}{2} = a t^{2} = x_{1}$
$k = \frac{y_{2} + y_{3}}{2} = 2 a t = y_{1}$

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Similar questions

P (x_{1}, y_{1})

y^{2} = 4 a x

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

Q

R,

Q R

The equation of the tangent at $(x_{1}, y_{1})$ to a curve $y^{2} = 4 a x$ at a point $(x_{1}, y_{1})$ is given by $y y_{1} = 2 a (x + x_{1})$

If the tangents to the parabola $y_{2} = 4 a x$ at the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ meet at the point $(x_{3}, y_{3})$ , then

y^{2} = 4 a x

(x_{1}, y_{1}), (x_{2}, y_{2})

(x_{3}, y_{3}),

y^{2} = 4 a x

(x_{1}, y_{1}), (x_{2}, y_{2})

(x_{3}, y_{3})

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