The point of intersection of tangents at $t_{1}$ and $t_{2}$ to the parabola $y^{2} = 4 a x$ is:

(a (t_{1} + t_{2}), a t_{1} t_{2})

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(a t_{1} t_{2}, a (t_{1} + t_{2}))

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(a t^{2}, 2 a t)

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(a t_{1} t_{2}, a (t_{1} - t_{2}))

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Solution

The correct option is D $(a t_{1} t_{2}, a (t_{1} + t_{2}))$
Fact: Equation of tangent to the parabola $y^{2} = 4 a x$ at $t$ is given by, $t y = x + a t^{2}$

Thus equation of tangent to the parabola at $t_{1}$ and $t_{2}$ are given by,
$t_{1} y = x + a t_{1}^{2}$ .......(1)
and $t_{2} y = x + a t_{2}^{2}$ .......(2)
Subtract (2) from (1)
$t_{1} y - t_{2} y = a t_{1}^{2} - a t_{2}^{2}$
$\Rightarrow y (t_{1} - t_{2}) = a (t_{1} + t_{2}) (t_{1} - t_{2})$
$\Rightarrow y = a (t_{1} + t_{2})$
Substitute value of y in (1)
$\Rightarrow a t_{1} (t_{1} + t_{2}) = x + a t_{1}^{2}$
$\Rightarrow x = a t_{1} t_{2}$

Hence the point of intersection is $(a t_{1} t_{2}, a (t_{1} + t_{2}))$

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Similar questions

If the tangents at P(t1) and Q(t2) in the parabola $y^{2} = 4 a x$ meet in T, then find the coordinates of point T.

A (a t_{1} t_{2}, a (t_{1} + t_{2})), B (a t_{2} t_{3}, a (t_{2} + t_{3}))

C (a t_{3} t_{1}, a (t_{3} + t_{1}))

A B C

t_{1} \neq t_{2} \neq t_{3})

[a t_{1} t_{2}, a (t_{1} + t_{2})], [a t_{2} t_{3}, a (t_{2} + t_{3})], [a t_{3} t_{1}, a (t_{3} + t_{1})]

t_{1} + t_{2} + t_{3} = - t_{1} t_{2} t_{3}

(a t_{1} t_{2}, a (t_{1} + t_{2}))

(a t_{2} t_{3}, a (t_{2} + t_{3}))

(a t_{3} t_{1}, a (t_{3} + t_{1}))

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

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

- 1 \leq t \leq 1

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