# Choose the correct answer in the point on the curve $$x^2 = 2y$$ which is nearest to the point $$(0, 5)$$ is

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

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.
edited Aug 30, 2013