# 18) For the curve $y = 4x ^ 3 - 2x ^5$, find all the points at which the tangent passes through the origin.

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is $y = 4x ^ 3 - 2x ^5$
Slope of tangent =

$\frac{dy}{dx} = 12x^2 - 10x^4$
Now, equation of tangent is
$Y-y= m(X-x)$
at (0,0)  Y =  0 and X = 0
$-y= (12x^3-10x^4)(-x)$
$y= 12x^3-10x^5$
and we have $y = 4x ^ 3 - 2x ^5$
$4x^3-2x^5= 12x^3-10x^5$
$8x^5 - 8x^3=0\\ 8x^3(x^2-1)=0\\ x=0\ \ \ \ \ \ and \ \ \ \ \ \ \ x = \pm1$
Now, when x = 0,

$y = 4(0) ^ 3 - 2(0) ^5 = 0$
when x = 1 ,

$y = 4(1) ^ 3 - 2(1) ^5 = 4-2=2$
when x= -1 ,

$y = 4(-1) ^ 3 - 2(-1) ^5 = -4-(-2)=-4+2=-2$
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

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