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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the point on the curve \( x^2 = 2y\) which is nearest to the point \((0, 5)\) is

\[ (A)\; (2 \sqrt2,4) \quad (B)\; (2 \sqrt2,0) \quad (C)\; (0, 0) \quad (D)\; (2, 2)\]

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Let $P(x,y)$ be a point on the curve.
The other point is $A(0,5)$
$z=PA^2=(x-0)^2+(y-5)^2$
$z=x^2+(y-5)^2$
Given $x^2=2y$
$z=2y+[y^2+25-10y]$
$\;\;=2y+y^2+25-10y$
$\;\;=y^2-8y+25$
Step 2:
$\large\frac{dZ}{dy}$$=2y-8$ [Differentiating with respect to y]
$\large\frac{d^2z}{dy^2}$$=2\Rightarrow$ +ve.
$\large\frac{dz}{dy}$$=0$
$2y-8=0$
$2y=8$
$y=4$
Step 3:
$\large\frac{d^2z}{dy^2}$$=+ve$ z is minimum
$x^2=2y$
$x^2=2\times 4$
$x^2=8$
$x=2\sqrt 2$
$\Rightarrow z$ is minimum at $(2\sqrt 2,4)$
$\Rightarrow \sqrt z$ is minimum at $(2\sqrt 2,4)$
Hence $A$ is the correct answer.
answered Aug 8, 2013 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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