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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the normal at the point \((1,1)\) on the curve \(2y + x^2 = 3\) is

\[ \begin{array}{l \qquad l \qquad l} (A)\; x + y = 0 \quad &(B) \;x – y = 0 \\(C)\; x + y +1 = 0 \quad & (D)\; x – y = 0\end{array} \]

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
The equation of the curve $2y+x^2=3$
Differentiating with respect to $x$
$2\large\frac{dy}{dx}$$+2x=0$
$2\large\frac{dy}{dx}$$=-2x$
$\large\frac{dy}{dx}$$=-x$
$\large\frac{dy}{dx}$ at $(1,1)=-1$=Slope of tangent.
Step 2:
Since slope of normal $=\large\frac{1}{slope \;of\; tangent}$$=1$
The equation of the normal is
$y-y_1$=Slope of normal$(x-x_1)$
$y-1=1.(x-1)$ or $y-1=x-1$
$x-y=0$
Part (B) is the correct answer.
answered Aug 13, 2013 by sreemathi.v
 

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