The coordinates of the point P(x,y) lying in the first quadrant on the  ellipse x2/8+y2/18=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 (8cos⁡θ,18sin⁡θ) Now  2x8+218ydydx=0⇒dydx=−9x4y⇒dydx(8cos⁡θ,18sin⁡θ)=−98cos⁡θ418sin⁡θ=     −92cotθ Hence the equation of the tangent at (8cos⁡θ,18sin⁡θ) is y   −18 sinθ  =  92 cotθ (x−8  cosθ)Therefore, the tangent cuts the coordinate axes at the points0,  18sinθ    and    8cosθ, 0Thus the area of the triangle formed by this tangent and the  coordinate axes is A=1218⋅8⋅1cos⁡θsin⁡θ=6cos⁡θsin⁡θ=12cosec⁡2θ But cosec 2θ is smallest when θ=π/4. Therefore A is  smallest when θ=π/4Hence the required point is 8⋅12⋅18⋅12=(2,3)