Question

For the curve y=4x^3 - 2x^5, find all points on the curve at which tangent passes through the origin
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Answer 1

Given :  curve y=4x^3 - 2x^5,

To find : points on the curve at which tangent passes through the origin

Solution:

y = 4x³ - 2x⁵

=> dy/dx  =  12x²  - 10x⁴

Slope of tangent  = 12x²  - 10x⁴

A line which passes through origin

y = m x

where m is the slope   m =  12x²  - 10x⁴

at point of tangent

y  =   ( 12x²  - 10x⁴ ) x    ( from line )

y = 4x³ - 2x⁵    ( from Curve )

Equating both

=> 4x³ - 2x⁵  =  12x³ -10x⁵

=> 8x⁵ - 8x³ = 0

=> 8x³ (x² - 1) = 0

=> x = 0 ,  ± 1

x = 0  , y = 0

x = 1 , y =  2

x = - 1  y = - 2

(0 , 0) , ( 1, 2) , ( - 1, - 2)  are the all points on curve at which tangent passes through the origin

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