Q98 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
A tangent to the curve $9 b^2 x^2-4 a^2 y^2=36 a^2 b^2$ makes intercepts of unit length on each of the coordinate axes, then the point $(a, b)$ lies on
- A$x^2-y^2=1$
- B$x^2+y^2=1$
- Correct answerC$4 x^2-9 y^2=1$
- D$4 x^2+9 y^2=1$
Solution
Equation of given curve is,
$$
\frac{x^2}{4 a^2}-\frac{y^2}{9 b^2}=1
$$
Let at point $(2 a \sec \theta, 3 b \tan \theta)$ on the curve (i).
So, equation of tangent at point is
$$
\frac{x}{\frac{2 a}{\sec \theta}}+\frac{y}{-\frac{3 b}{\tan \theta}}=1
$$
According to the question,
$$
2 a=\sec \theta \text { and } 3 b=-\tan \theta
$$
So,
$$
4 a^2-9 b^2=1
$$
On taking locus of point $(a, b)$, we are getting
$$
4 x^2-9 y^2=1 \text {. }
$$
Hence, the correct answer is option: 'C'.