Question

Tangent at P(2,8) on the curve `y=x^(3)` meets the curve again at Q.
Find coordinates of Q.

Answer 1

Equation of tangent at (2,8) is y =12x -16

Solving this with `y=x^(3)`
`x^(3) -12x+16=0`
This cubic will give all points of intersection of line and curve `y=x^(3)` i.e., point P and Q . (see figure)
But since line is tangent at P so x=2 will be a repeated root of equation `x^(3) -12x +16=0` and another root will be x=h. Using theory or equations:
`"sum of roots"" "rArr " "2+2+h=0" "rArr" "h=-4`
Hence coordinates of Q are (-4,-64)

Comments

I didn't get why (x=2) will be a repeated root of the equation x³-12x+16

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It is concept of tangent that if two intersection point coincides then that line is tangent to that point.

Therefore, we considered x = 2 will be a repeated root of equation. Although we can find this result algebraically that x = 2 is repeated root of equation

x^3 - 12x + 16 = 0
⇒ (x - 2) (x^2 + 2x - 8) = 0
⇒ (x - 2) (x - 2) (x + 4) = 0
⇒ x = 2, 2, -4

Hence, x = -2 is repeated root and other root is x = -4.