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

- (2, 3)
- (√8, 0)
- (3,2)
- (√18, 0)

Solution

The correct option is **A** (2, 3)

Any point on the ellipse is given by (√8cosθ,√18sinθ) Now 2x8+218ydydx=0⇒dydx=−9x4y

⇒dydx|(√8cosθ,√18sinθ)=−9√8cosθ4√18sinθ=−√92cotθ.

Hence the equation of the tangent at (√18cosθ,√18sinθ) is

y - √18sinθ=−√92cotθ(x−√8cosθ)

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 = 12√18√8 . 1cosθsinθ=6cosθsinθ=12cosec2θ.

But cosec2θ is smallest when θ=π4. Therefore A is smallest when θ=π4.

Hence the required point is (√8.1√2,√18.1√2) = (2, 3)

Any point on the ellipse is given by (√8cosθ,√18sinθ) Now 2x8+218ydydx=0⇒dydx=−9x4y

⇒dydx|(√8cosθ,√18sinθ)=−9√8cosθ4√18sinθ=−√92cotθ.

Hence the equation of the tangent at (√18cosθ,√18sinθ) is

y - √18sinθ=−√92cotθ(x−√8cosθ)

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 = 12√18√8 . 1cosθsinθ=6cosθsinθ=12cosec2θ.

But cosec2θ is smallest when θ=π4. Therefore A is smallest when θ=π4.

Hence the required point is (√8.1√2,√18.1√2) = (2, 3)

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