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प्रश्न
Find points on the curve `x^2/9 + "y"^2/16 = 1` at which the tangent is parallel to x-axis.
उत्तर
The equation of the given curve is `x^2/9 + "y"^2/6 = 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 x-axis if the slope of the tangent is i.e., `0 (- 16 x)/"9y" = 0,` which is possible if x = 0.
Then, `x^2/9 + "y"^2/6 = 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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