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