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Find points on the curve yx29+y26 = 1 at which the tangent is parallel to y-axis. - Mathematics

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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. 

Sum
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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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Chapter 6: Application of Derivatives - Exercise 6.3 [Page 212]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 6 Application of Derivatives
Exercise 6.3 | Q 13.2 | Page 212

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