The coordinate of the point p(x,y)  lying in the first quadrant on the ellipsex28+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

Any point on the ellipse is given by (8cosθ,18sinθ) Now      2x8+218ydydx=0⇒dydx=−9x4y ⇒dydx|(8cosθ,18sinθ)=−98cosθ418sinθ=−92cotθ Hence the equation of the tangent at (8cosθ,18sinθ)  isy−18sinθ=−92cotθ(x−8cosθ) Therefore , the tangent cuts the coordinate axes at the points(0,18sinθ)  and (8cosθ,0) Thus the area of the triangle formed by this tangent and the coordinate axes isA=1218.8.1cosθsinθ      =6cosθsinθ=12cosec  2θ But cosec  2θ  is smallest when θ=π/4. Therefore A is smallest when θ=π/4. Hence the required point is (8.12,18.12)=(2,3)