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Question

The coordinates of the point P(x,y) lying in the first quadrant on the ellipse x28 + y218 = 1 so that the area of the triangle formed by the tangent at P and the coordinate axes is the smallest, are given by  
  1. (2, 3)
  2. (8, 0)
  3. (3,2)
  4. (18, 0)


Solution

The correct option is A (2, 3)
Any point on the ellipse is given by (8cosθ,18sinθ)  Now  2x8+218ydydx=0dydx=9x4y
dydx|(8cosθ,18sinθ)=98cosθ418sinθ=92cotθ.
Hence the equation of the tangent at (18cosθ,18sinθ) is
y - 18sinθ=92cotθ(x8cosθ) 
Therefore, the tangent cuts the coordinates axes at the points (0,18sinθ) and (18cosθ,0) 
Thus the area of the triangle formed by this tangent and the coordinate axes is        A = 12188 . 1cosθsinθ=6cosθsinθ=12cosec2θ.
But cosec2θ is smallest when θ=π4. Therefore A is smallest when θ=π4
Hence the required point is (8.12,18.12) = (2, 3)

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