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Tangent To a Parabola

Tangents draw...

Question

Tangents drawn at A $(a t_{1}^{2}, 2 a t_{1})$ and B $(a t_{2}^{2}, 2 a t_{2})$ intersect at (p,q).

Then p and q are and of corresponding x and y coordinates of A and B respectively $(g i v e n t_{1} \neq t_{2})$

GM, HM

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GM, AM

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AM, GM

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AM, HM

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Solution

The correct option is B
GM, AM

let's write equation of tangents at the points A nad B. on solving these tangents we get the point of intersection of these lines.

Tangent at A

$t_{1} y = x + a t_{1}^{2}$ .......(1)

Tangent at B

$t_{2} y = x + a t_{2}^{2}$ .......(1)

solving (1) and (2)

(1)-(2)

$y (t_{1} - t_{2}) = a (t_{1}^{2} - t_{2}^{2})$

= $a (t_{1} - t_{2}) (t_{1} + t_{2})$

$y = a (t_{1} + t_{2})$

$(s i n c e t_{1} - t_{2} \neq 0)$

= $\frac{2 a t_{1} + 2 a t_{2}}{2}$

= AM of x coordinates

Solving (3)&(1)

$x = t_{1} y - a t_{1}^{2}$

$= t_{1} . a . (t_{1} + t_{2}) - a t_{1}^{2}$

$= a t_{1}^{2} + a t_{1} t_{2} - a t_{1}^{2}$

$= a t_{1} t_{2}$

$\sqrt{a t_{1}^{2} \times a t_{2}^{2}}$

=GM of y coordinates

Hence option (b) is correct

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