The point of intersection of the tangents to the parabola $y^{2} = 4 x$ at the points where parameter 't' has value 3 and 5.

(10, 6)

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(4, 9)

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(15, 8)

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(15, 10)

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Solution

The correct option is C
(15, 8)

If we take the tangent equations at given points and solve them we get the point of intersection. Its better if we memorize this point given by,

P $\equiv (a t_{1} t_{2}, a (t_{1} + t_{2}))$

Here $t_{1} = 3; t_{2} = 5; a = 1$

$∴ P \equiv (15, 8)$

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Similar questions

The point of intersection of the tangents to the parabola $y^{2} = 4 x$ at the points where parameter 't' has value 3 and 5.

The point of intersection of the tengents to the parabola $y^{2} = 4 x$ at the points, where the parameter 't' has the value 1 and 2, is

P (t),

t

y^{2} = 4 a x

y^{2} = 4 x

y^{2} = - 8 (x - a)

A

C

O (0, 0), A, B (a, 0), C

y^{2} = 4 x

A

C

D

y^{2} = - 8 (x - a)

E

D A E C

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