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Question

A line L is perpendicular to the curve y=x24−2 at its point P and passes through (10, -1). The coordinates of the point P are

A
(2, -1)
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B
(6, 7)
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C
(0, -2)
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D
(4, 2)
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Solution

The correct option is D (4, 2)
Let point (x1,y1) on curve y=x24−2
Slope of tangent is dydx=x12
So slope of perpendicular line is −2x1 ...(1)
That perpendicular line also passes through (x1,y1) and (10,−1)
Slope of line from these point is y1+1x1−10 ...(2)
Equating (1) and (2) and solving we get,
x1=4 and y1=2
Thus point is (4,2)
Hence, option 'D' is correct.

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