AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

The equation of given curve is,

x2

4a2

−

y2

9b2

=1 ......(I)
Let at point (2a secθ ,3b tanθ ) on the curve in equation (I),
So, equation of tangent at point is,

x

2a sec‌θ

+

y

3b tan‌θ

=1
As per the question,
2a=sec‌θ‌and‌3‌b=−tan‌θ
So,
4a2−9b2=1
Take locus of point (a,b) then,
4x2−9y2=1