Question

at which point on the ellipse x^2/8 +y^2/18=1 must a tangent be drawn such that the area of the triangle formed by the tangent and the coordinate axes is smallest

Answer 1

Let the point be P(x1, y1).

Ellipse :  x^2 / 8 + y^2 / 18 = 1

Equation of the tangent :   x x1 /8 + y y1 / 18 = 1

The x and y intercepts of the tangent are = 8/x1  and 18/y1 respectively.

A = Area of the triangle = 1/2 * 8/x1 * 18/y1 = 72 /(x1 y1)
A^2 = 72^2 / (x1^2 y1^2)

If the Area A is to be minimum possible, then x1^2 y1^2 must be the maximum.

y1^2 = 18 (8 - x1^2) / 8

So  F =  x1^2 (8 - x1^2) must be maximum.

Differential: dF/ dx1 = 2 x1 (8 -x1^2) + x1^2 (-2 x1) = 0
 => 8 - x1^2 - x1^2 = 0

 =>  x1 = +2 or - 2
 =>  y1 = +3 or -3