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Find the point on the curve y = x3 – 11x + 5 at which the tangent has the equation y = x – 11.
Asked by Topperlearning User | 11 Aug, 2014, 10:06: AM

Let the required point be P(x1, y1), since (x1, y1) lies on y = x3 – 11x + 5.

y1 =   …(i)
Now, y = x3 – 11x + 5

Since the line y = x – 11 is tangent at the point (x1, y1). Therefore,
Slope of the tangent at (x1, y1) = (Slope of the line y = x – 11).
(Slope of the line x – y – 11 = 0)
Now, [Using (i)]
[Using (i)]
So, two points are (2, –9) and (–2, 19). Of these two points (–2, 19) does not lie on y = x – 11. Therefore, the required point is (2, – 9).
Answered by | 11 Aug, 2014, 12:06: PM

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