Question

Given that the equation of the tangent and normal to the ellipse (x^2/a^2) + (y^2/b^2) = 1 at the point P (acosu , bsinu) are respectively
bxcosu + aysinu = ab and
axsinu + bycosu = (a^2 - b^2)sinucosu.
Given that the tangents cuts the x - axis at A and the y - axis at B and that the normal cuts the x - axis at C and the y - axis at D. Show that as u varies, the locus of the midpoint of CD is
4[(a^2)*(x^2)]+ 4[(b^2)*(y^2)] = (a^2 - b^2)^2
Given that a = 5 and b = 4, sketch this locus

Answer 1

Point C: $y=0\implies ax=(a^2-b^2)cosu$

Point D: $x=0\implies by=(a^2-b^2)sinu$

Midpoint of CD: $x_M=\frac{(a^2-b^2)cosu}{2a}, y_M=\frac{(a^2-b^2)sinu}{2b}$

$cos^2u+sin^2u=1$

$\frac{4a^2x_M^2}{(a^2-b^2)^2}+\frac{4b^2y_M^2}{(a^2-b^2)^2}=1$

$4a^2x_M^2+4b^2y_M^2=(a^2-b^2)^2$