Question

Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}$=1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis

Answer 1

The equation of the given curve is $\frac{x^2}{9}+\frac{y^2}{16}$=1.

On differentiating both sides with respect to x, we have:

$\frac{2x}{9}+\frac{2y}{16}$.$\frac{dy}{dx}$=0

=$\frac{dy}{dx}$=$\frac{-16x}{9y}$

(i) The tangent is parallel to the x-axis if the slope of the tangent is i.e., 0 $\frac{-16x}{9y}$=0,

which is possible if x = 0.

Then,$\frac{x^2}{9}+\frac{y^2}{16}$=1 for x=0

y2=16 ⇒ y±4

Hence, the points at which the tangents are parallel to the x-axis are (0, 4) and (0, − 4).

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(ii) The tangent is parallel to the y-axis if the slope of the normal is 0, which

gives $\frac{-1}{(\frac{-16x}{9y})}$=$\frac{9y}{16x}$=0
⇒ y=0.

Then, $\frac{x^2}{9}+\frac{y^2}{16}$ =1 for y=0.

x=±3.

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