Question

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

Answer 1

Equation of the curve is given as,

y=4 x 3 −2 x 5 (1)

The slope of the tangent to the curve is given as,

slope=( dy dx )

Hence, the slope of the tangent of the given curve is,

dy dx = d( 4 x 3 −2 x 5 ) dx =12 x 2 −10 x 4

Equation of the tangent at point ( x 0 , y 0 ) is given as,

y− y 0 =m( x− x 0 )

Hence, equation of the tangent passing through ( 0,0 ) with slope 12 x 2 −10 x 4 is given by,

y−0=( 12 x 2 −10 x 4 )( x−0 ) y=12 x 3 −10 x 5

Compare the above equation with equation (1),

4 x 3 −2 x 5 =12 x 3 −10 x 5 8 x 5 −8 x 3 =0 x 3 ( x 2 −1 )=0 x=0, ±1

Coordinate of y when x=0 is,

4 ( 0 ) 3 −2 ( 0 ) 5 =0

Coordinate of y when x=1 is,

4 ( 1 ) 3 −2 ( 1 ) 5 =2

Coordinate of y when x=−1 is,

4 ( −1 ) 3 −2 ( −1 ) 5 =−2

Thus, the points on the given curve at which tangent passes through origin are ( 0,0 ), ( 1,2 ) and ( −1,−2 ).