Question

For the curve y = 4x^3 − 2x^5 , find all the points at which the tangents passes through the origin.

Answer 1

HELLO DEAR,

The equation of the given curve is
y = 4x³ - 2x^5.-------( 1 )

On differentiating , w.r.t. x,
we get,
dy/dx = 12x² - 10x⁴.

let the required point be p(x , y).
the equation of tangent to the given curve at p(x , y) is given by

Y - y = (12x² - 10x⁴)(X - x).
if it passes through (0,0) then,
we have
y = (12x² - 10x⁴)x
4x³ - 2x^5 = 12x³ - 10x^5 ----- from-----( 1 )
8x³ - 8x^5 = 0
8x³(1 - x²) = 0
x = 0 , x = 1 , x = -1
put x = 0 in ( 1 ), we get y = 0.
put x = 1 , in ( 1 ) , we get y = 2.
put x = -1 in ( 1 ) ,we get y = -2.
So, the required points are (0,0), (1,2) (-1, -2).

I HOPE ITS HELP YOU DEAR,
THANKS