Question

The point on the curve $y=x^2$ which is nearest to (3, 0) is

• A. (1, -1)
• B. (-1, 1)
• C. (-1, -1)
• D. (1, 1)

Answer 1

The correct option is D (1, 1)

Let P(x,y) be a point on the parabola $x^2=y$
Distance between P(x,y) and (3,0) is
$S^2=(x-3{)}^2+(y-0{)}^2$
$S^2=x^2-6x+9+x^4$
$2S\frac{dS}{dx}=2x-6+4x^3$
For maxima or minima,
$\frac{dS}{dx}=0$
$\Rightarrow 4x^3+2x-6=0$
$\Rightarrow x=1$
$f^{\prime \prime }\left(x\right)>0$ at $x=1$
$\Rightarrow y=1$
Hence, the nearest point is (1,1).