Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • 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,
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,
$\Rightarrow \large\frac{y}{4}=\frac{x-2}{4-2}$
$\Rightarrow \large\frac{y}{4}=\frac{x-2}{2}$
$\Rightarrow 2y=4x-8$
Dividing throughout by 2 we get,
$\Rightarrow 2x-y-4=0$
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.
$\Rightarrow x=3$
When $x=3$,
Hence the required point is $(3,1)$
answered Jul 10, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App