Question

The points on the curve `"x"^2/9 + "y"^2/16` = 1 at which the tangents are parallel to the y-axis are:

Options

• (0, ±4)

• (±4, 0)

• (±3, 0)

• (0, ±3)

Answer 1

(±3, 0)

Explanation:

`"x"^2/9 + "y"^2/16` = 1

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

The slope of normal at any point (x, y) to the curve = `(-"dx")/("dy") = (9"y")/(16"x")`

As a tangent to the curve at the point (x, y) is parallel to the y-axis

`(9"y")/(16"x")` = 0

y = 0 and x = ±3

∴ Points = (±3, 0)