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  • Find the points of intersection of the tangents at the ends of the latus rectum to the parabola .<div clas

Find the points of intersection of the tangents at the ends of the latus rectum to the parabola x^2 = 4y.

Option: 1

(0,2)


Option: 2

(0,1)


Option: 3

(2,0)


Option: 4

(1,0)


Answers (1)

best_answer

 

 

Point of Intersection of Tangent -

Point of Intersection of Tangent

\\ {\text { Two points, } P \equiv\left(a t_{1}^{2}, 2 a t_{1}\right) \text { and } Q \equiv\left(a t_{2}^{2}, 2 a t\right) \text { on the parabola } y^{2}=4 a x .} \\ {\text { Then, equation of tangents at } P \text { and } Q \text { are }} \\ {t_{1} y=x+a t_{1}^{2}} \\ {t_{2} y=x+a t_{1}^{2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots() \\ {\text { solving (i) at }_{2}^{2}} \\ {\text { we get, } x=a t_{1} t_{2}, y=a\left(t_{1}+t_{2}\right)} \\ {\text { Point of Intersection of tangents drawn at point } P \text { and } Q \text { is }} \\ {\left(a t_{1} t_{2}, a\left(t_{1}+t_{2}\right)\right)}

Point of Intersection of tangents drawn at point P and Q is 

-

 

 

The equation of the given parabola is x^2 = 4y.

We have, 4a = 4 or, a = 1

Let the end-points of the latus rectum are L(2a, a) and L'(-2a, a).

Therefore L = (2, 1) and L'= (-2, 1).

As we know that the point of intersection to the tangents at

(2at_1,at_1^2) and (2at_2,at_2^2) to the parabola x^2=4ay is (a(t_1+t_2),at_1t_2)

Thus, the point of intersection of the tangents at L(2,1) and L'(-2, 1) is (0,1).

 

Posted by

Anam Khan

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