Consider the parabola y2=8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum
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 Δ1/Δ2 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
Δ1/Δ2=2
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