Q4 of 149 Page 16

Find a point on the curve y = x3 – 3x where the tangent is parallel to the chord joining (1, – 2) and (2, 2).

Given:


The curve y = x3 – 3x


First, we will find the Slope of the tangent


y = x3 – 3x


(x3) – (3x)


⇒ = 3x3 – 1 – )


⇒ = 3x2 – 3 ...(1)


The equation of line passing through (x0,y0) and The Slope m is y – y0 = m(x – x0).


so The Slope, m =


The Slope of the chord joining (1, – 2) & (2,2)


⇒


⇒


⇒ = 4 ...(2)


From (1) & (2)


3x2 – 3 = 4


⇒ 3x2 = 7


⇒ x2 =


⇒ x =


y = x3 – 3x


⇒ y = x(x2 – 3)


⇒ y = (()2 – 3)


⇒ y = (( – 3)


⇒ y = ()



⇒ y = ()


Thus, the required point is x = & y = ()


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