Answer:
Given
curve is
y
= x3 ? 11x + 5 ?.(1)
Let
P(h, k) be a point on (1)
K = h2
? 11 x + 5 ?. (2)
(1)
Given
that slope of tangent to (1) = 1
When
h = 2,
(2)
K = 8 ? 22
+ 5 = ?9
When
h = ?2,
(2)
K = ?8 +
22 + 5 = 19
Point on
(1) are (2, ?9) and (?2, 19)
Now
equation of tangent to curve (1)
At
(2, ?9) y ? k = m (x ? h)
y + 9 = 1
(x ? 2)
y = x ? 11
At
(?2, 19)
y
? 19 = 1 (x + 2)
y = x +
21.
Hence
the point on (1) where the tangent to (1) is
y
= ?11, is (2, ?9).
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