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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the line \(y = x + 1\) is a tangent to the curve \(y^2 = 4x\) at the point

\[ (A)\; (1, \: 2) \quad (B) \;(2, \: 1) \quad (C) \;(1, \: -2) \quad (D)\; (-1, \: 2)\]

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1 Answer

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Toolbox:
  • If a tangent line to the curve $y=f(x)$ makes an angle $\theta$ with $x$-axis in the +ve direction,then $\large\frac{dy}{dx}$=slope of the tangent=$\tan \theta$
  • Slope of the normal is $\large\frac{-1}{\Large\frac{dy}{dx}}$
Step 1:
Equation of the curve is $y^2=4x$
Differentiating w.r.t $x$ we get,
$2y\large\frac{dy}{dx}$$=4$
Therefore $\large\frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}$
$\Rightarrow m=\large\frac{2}{y}$
Step 2:
Equation of the line is $y=x+1$----(1)
Slope of the given line is $1$.
$\Rightarrow \large\frac{2}{y}$$=1$
Therefore $y=2$
Step 3:
When $y=2$,substituting for $y$ in equ(1)
$2=x+1$
$\Rightarrow x=1$
Hence the required point is $(1,2)$
The correct option is $A$.
answered Jul 12, 2013 by sreemathi.v
 

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