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Question

# The area of the triangle formed by the tangent and the normal to the parabola y2=4ax, both drawn at the same end of the latus rectum, and the axis of the parabola is

A

22a2

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B

2a2

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C

4a2

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D

None of these

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Solution

## The correct option is C 4a2 One end of the latus rectum is P(a, 2a) The equation of the tangent PT at P(a,2a) is y.2a = 2a(x+a), i.e., y =x+a. The equation of normal PN at (a, 2a) is y + x =2a+a, i.e., y + x =3a. Solving y=0 and y=x+a, we get x=-a, y=0. Solving y=0, y+x=3a, we get x=3a, y=0. The area of the triangle with vertices P(a, 2a), T(-a, 0), N(3a, 0) is 4a2.

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