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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the normal to the curve \(x^2 = 4y\) passing \((1,2)\) is

\[ \begin{array}{l \qquad l \qquad l} (A)\; x + y = 3 \quad &(B)\; x -y = 3 \\(C)\; x + y = 1 \quad & (D)\; x - y = 1\end{array} \]

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • Slope of the normal $=-\large\frac{1}{Slope\;of\;tangent}$
Step 1:
Given :
$x^2=4y$
Differentiating with respect to $x$
$2x=4\large\frac{dy}{dx}$
$\large\frac{dy}{dx}=\frac{2x}{4}=\frac{x}{2}$
$\large\frac{x}{2}$=Slope of the tangent.
Step 2:
Slope of the normal $=-\large\frac{1}{Slope\;of\;tangent}$
$\qquad\qquad\qquad\quad=\large\frac{-1}{x/2}$
$\qquad\qquad\qquad\quad=\large\frac{-2}{x}$
$\qquad\qquad\qquad\quad=\large\frac{-2}{x_1}$ (at $x_1,y_1)$
Step 3:
Equation of normal $y-y_1=\large\frac{-2}{x_1}$$(x-x_1)$------(1)
It passes through $(1,2)$
$2-y_1=\large\frac{-2}{x_1}$$(1-x_1)$
$2x_1-x_1y_1=-2+2x_1$
$x_1y_1=2$----(2)
$y_1=\large\frac{2}{x_1}$------(3)
Step 4:
The point $(x_1,y_1)$ lies on $x^2=4y$
From (2) & (3)
$x_1^2=4.\large\frac{2}{x_1}$
$x_1^3=8$
$x_1=2$
From (3) $4=4y_1$
$y_1=1$
Step 5:
Putting these values in (1) equation of normal is
$y-1=\large\frac{-2}{2}$$(x-2)$
$y-1=-x+2$
$x+y=3$
Part (A) is the correct answer.
answered Aug 13, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

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