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Question Number 162253 by ZiYangLee last updated on 28/Dec/21

The tangent of a parabola y^{2} = 4 a x at the point

P ({a p}^{2}, 2 a p) intersects the line x + a = 0 at T .

(i) If M is the midpoint of P T, find the

coordinates of M in terms of a and p .

(ii) Prove that the equation of locus of M is

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

Answered by som(math1967) last updated on 28/Dec/21

y^{2} = 4 a x

2 y \frac{d y}{d x} = 4 a

{[\frac{d y}{d x}]}_{{a p}^{2}, 2 a p} = \frac{4 a}{4 a p} = \frac{1}{p}

∴ s l o p e o f t a n j e n t = \frac{1}{p}

e q u a t i o n o f t a n j e n t

(y - 2 a p) = \frac{1}{p} (x - {a p}^{2})

y p - 2 {a p}^{2} = x - {a p}^{2}

x - y p + {a p}^{2} = 0

p u t x = - a [t o f i n d p t T]

- a + {a p}^{2} = y p

y = \frac{{a p}^{2} - a}{p}

∴ P ({a p}^{2}, 2 a p) T (- a, \frac{{a p}^{2} - a}{p})

∴ m i d p t o f P T

x = \frac{- a + {a p}^{2}}{2} y = \frac{2 a p + \frac{{a p}^{2} - a}{p}}{2} = \frac{3 {a p}^{2} - a}{2 p}

a n s i)

x = \frac{- a + {a p}^{2}}{2} \Rightarrow p^{2} = \frac{2 x + a}{a}

y = \frac{3 {a p}^{2} - a}{2 p}

2 y p = 3 {a p}^{2} - a

4 y^{2} p^{2} = {(3 {a p}^{2} - a)}^{2}

4 y^{2} (\frac{2 x + a}{a}) = {(\frac{6 a x + 3 a^{2} - a^{2}}{a})}^{2}

4 y^{2} (2 x + a) = {(6 a x + 2 a^{2})}^{2} \times \frac{a}{a^{2}}

4 y^{2} (2 x + a) = \frac{4 a^{2} {(3 x + a)}^{2}}{a}

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

y^{2} (2 x + a) = a {(3 x + a)}^{2} [p r o v e d]

Commented by peter frank last updated on 29/Dec/21

great sir