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).