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Perpendicular Distance of a Point from a Line

The point of ...

Question

The point of intersection of tangents at the points on the parabola $y^{2} = 4 x$ whose ordinates are $4$ and $6$ is

A

(6, 5)

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B

(7, 3)

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C

(9, 10)

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D

(6, 10)

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Solution

The correct option is A $(6, 5)$
The point of intersection of tangents at any two points $A (a t_{1}^{2}, 2 a t_{1}) and B (a t_{2}^{2}, 2 a t_{2})$ on the parabola $y^{2} = 4 a x$ is given by $(a t_{1} t_{2}, a (t_{1} + t_{2}))$
$2 a t_{1} = 4, 2 a t_{2} = 6$
For the parabola $y^{2} = 4 x$
$\Rightarrow a = 1 \Rightarrow t_{1} = 2, t_{2} = 3$
Then point of intersection of tangents is $(1 \times 2 \times 3, 1 \times (2 + 3)) = (6, 5)$

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Perpendicular Distance of a Point from a Line

Standard XII Mathematics

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