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

The correct answer is Option (4) → No such points exist.

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 x-axis, its slope = 0

$⇒\frac{16x_1}{9y_1}=0⇒16x_1=0⇒x_1=0$

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

$16.0^2-9{y_1}^2 = 144⇒ y^2=-16 ⇒ y = ±4i$.

Thus, there is no real point at which the tangent is parallel to x-axis. In other words, the given curve has no tangent parallel to x-axis.