Question

Find points on the curve `x^2/9 + "y"^2/16` = 1 at which the tangent is parallel to y-axis. 

Answer 1

The equation of the given curve is `x^2/9 + "y"^2/16 = 1`.

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

`(2x)/9 + (2"y")/16 * "dy"/"dx" = 0`

`=> "dy"/"dx" = (- 16 x)/(9"y")`

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

`(-1)/(((- 16x)/"9y")) = "9y"/(16x) = 0` ⇒ y = 0

Then, `x^2/9 + "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).