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Consider the parabola y2=8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum and the point P(1/2,2) on the parabola, and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then Δ12 is

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Solution:

The extremes points of the latus rectum are: (2,4)and(2,−4)

Hence the area of triangle formed by these points and P is

Δ1=6 (directly apply the formula)

Equation of tangent at (2,4) is

y=x+2⋯(i)

Equation of tangent at (2,−4) is

−y=x+2⋯(ii)

Equation of tangent at (1/2,2) is

y=2x+1⋯(iii)

Point of intersection of (i)and(ii) is (1,3) and of (ii)and(iii) is (−1,−1)

again by applying the formula for area of a triangle we find that

Δ2=3

hence

Δ12=2

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