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प्रश्न
The points on the curve `"x"^2/9 + "y"^2/16` = 1 at which the tangents are parallel to the y-axis are:
विकल्प
(0, ±4)
(±4, 0)
(±3, 0)
(0, ±3)
MCQ
उत्तर
(±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)
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