Find the points on the curve (x^2)/9-(y^2)/16= 1 at which the tangent…

The correct answer is Option (2) → (3, 0) and (−3, 0)

Let $P(x_1,y_1)$ be a required point.

The given curve is $\frac{x^2}{9}-\frac{y^2}{16}= 1$ i.e. $16x^2 - 9y^2 = 144$   ...(1)

Diff. (1) w.r.t. x, we get

$16.2x-9.2y\frac{dy}{dx}= 0⇒ \frac{dy}{dx}=\frac{16x}{9y}⇒\left(\frac{dy}{dx}\right)_P=\frac{16x_1}{9y_1}$

Since the tangent at P is parallel to y-axis, $\left(\frac{dy}{dx}\right)_P=0$

$⇒\frac{9y_1}{16x_1}=0⇒9y_1=0⇒y_1=0$

As $P(x_1,y_1)$ lies on the curve (1), we get

$16{x_1}^2-9.0^2=144⇒{x_1}^2=9⇒x_1=±3$.

Hence, the points at which the tangents to the given curve are parallel to y-axis are (3, 0) and (−3, 0).