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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find a point on the curve $y = (x – 2)^2 $ at which the tangent is parallel to the chord joining the points \((2, 0)\) and \((4, 4).\)

$\begin{array}{1 1} (A)\;(1,3) \\(B)\;(-1,3) \\(C)\;(1,-3) \\ (D)\;(3,1) \end{array} $

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1 Answer

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Toolbox:
  • If $y=f(x)$,then $\big(\large\frac{dy}{dx}\big)_P$=slope of the tangent to $y=f(x)$ at the point $P$.
  • Equation of line with given two points is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}$
Step 1:
Given : $y=(x-2)^2$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=2(x-2)$
Equation of the chord is $\large\frac{y-y_1}{y_2-y_1}=\large\frac{x-x_1}{x_2-x_1}$
Here $(x_1,y_1)=(2,0)$ and $(x_2,y_2)=(4,4)$
Now substituting the values we get,
$\large\frac{y-0}{4-0}=\frac{x-2}{4-2}$
$\Rightarrow \large\frac{y}{4}=\frac{x-2}{4-2}$
$\Rightarrow \large\frac{y}{4}=\frac{x-2}{2}$
$\Rightarrow 2y=4x-8$
Dividing throughout by 2 we get,
$y=2x-4$
$\Rightarrow 2x-y-4=0$
Step 2:
Slope of the given chord is $\bigg(-\large\frac{coeff\;of \;x}{coeff\;of \;y}\bigg)$
Therefore slope of the chord is $\bigg(\large\frac{-2}{-1}\bigg)=\frac{2}{1}$
$\Rightarrow 2$
Since the tangent to the curve to the chord,their slopes will be equal.
(i.e)$2(x-2)=2$
$\Rightarrow x=3$
When $x=3$,
$y=(x-2)^2$
$y=(3-2)^2$
$y=1$
Hence the required point is $(3,1)$
answered Jul 10, 2013 by sreemathi.v
 

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