Question

The minimum distance of the curve $y^{2}=2x^{3}+9-3x^{2}$ from point $Q\left(\right.1,0\left.\right)$ is

• A. $2$
• B. $2\sqrt{2}$
• C. $4\sqrt{2}$
• D. $8$

Answer 1

Answer: (B)

$P(x, y), Q(1,0) $
$P Q=\sqrt{(x-1)^2+y^2} $
$P Q^2=x^2+1-2 x+y^2$
$P Q^2=x^2-2 x+1+2 x^3-3 x^2+9$
$P Q^2=2 x^3-2 x^2-2 x+10$
$\frac{d P Q^2}{d x}=6 x^2-4 x-2=0 $
$6 x^2-4 x-2=0$
$3 x^2-2 x-1=0 $
$3 x^2-3 x+x-1=0 $
$3 x(x-1)+(x-1)=0 $
$(x-1)(3 x+1)=0 $
$x=1, x=-\frac{1}{3} $
$\frac{d^2 P Q^2}{d x^2}=12 x-4$
$\text { So,PQ is minimum at } x=1 y^2=2+9-3 $
$=8 $
$(P Q)_{\text {minimum }}=\sqrt{0+8}=2 \sqrt{2}$