Differentiate wrt x: 2x/a^2+2ydy/dx/b^2=0. Therefore dy/dx=-(x/a^2)(b^2/y)=-xb^2/ya^2.
The equation of the tangent is given by y=mx+c where m=-Xb^2/Ya^2 at the point (X,Y) where the tangent line intercepts each axis equidistant from the origin. Since x^2/a^2+y^2/b^2=1 we can find Y when x=X: Y^2=b^2(1-X^2/a^2), so Y=bsqrt(1-X^2/a^2). In the equation y=mx+c, the y intercept is c (when x=0) and the x intercept is -c/m. These are equal when c=-c/m, so, if c not zero, m=-1, and -Xb^2/Ya^2=-1, Xb^2=Ya^2=absqrt(a^2-X^2), that is, Xb=asqrt(a^2-X^2); X^2b^2=a^2(a^2-X^2)=a^4-a^2X^2, so X^2=a^4/(a^2+b^2) and X=a^2/sqrt(a^2+b^2). Y=bsqrt(1-a^2/(a^2+b^2))=bsqrt(b^2/(a^2+b^2))=b^2/sqrt(a^2+b^2).
The point (X,Y) is therefore (a^2/sqrt(a^2+b^2),b^2/sqrt(a^2+b^2)).
[c=0 in the equation c=-c/m is not a solution, because the curve is an ellipse when a and b have non-zero values. The line y=mx (both intercepts are zero) passes through the origin (0,0) which cannot be the tangent line for any point on the ellipse.]