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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
+1 vote

For the curve \(y = 4x^3 – 2x^5\), find all the points at which the tangent passes through the origin.

$\begin{array}{1 1} (A)\;(0,0),(1,-2),(-1,2) \\ (B)\;(0,0),(-1,-2),(1,2)\\ (C)\;(0,0),(1,2),(-1,2) \\ (D)\;(0,0),(1,-2) \end{array} $

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1 Answer

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Let $(x_1,y_1)$ be the required point on the given curve $y=4x^3-2x^5$
$\Rightarrow y_1=4x_1^3-2x_1^5$-----(1)
Differentiating equ(1) w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=12x^2-10x^4$
Therefore $\large\frac{dy}{dx}_{(x_1,y_1)}$$=12x_1^2-10x_1^4$
Step 2:
Equation of the tangent is $y-y_1=(12x_1^2-10x_1^4)(x-x_1)$
This passes through the origin hence $0-y_1=(12x_1^2+10x_1^4)(0-x_1)$
$\Rightarrow y_1=12x_1^3-10x_1^5$------(2)
Subtract equ(2) from equ(1)
$0=-8x_1^3+8x_1^5$
$\Rightarrow 8x_1^3(x_1^2-1)=0$
$\Rightarrow x_1=0$ or $x_1=\pm 1$
Step 3:
When $x_1=0$ from equ(2) $y=0$
When $x_1=1$ from equ (2)
$y_1=4(1)^2-2(1)^5$
$\;\;\;\;=4-2$
$\;\;\;\;=2$
When $x_1=-1$ from equ(2)
$y_1=4(-1)^3-2(-1)^5$
$\;\;\;\;=-4-(-2)$
$\;\;\;\;=-2$
Step 4:
Hence the required points are $(0,0),(1,2)$ and $(-1,-2)$
answered Jul 12, 2013 by sreemathi.v
 

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