The minimum distance of the curve y2=2x3+9-3x2 from point Q(.1,0.) is

Question

The minimum distance of the curve y2=2x3+9-3x2 from point Q(.1,0.) is (A) 2 (B) 2√2 (C) 4√2 (D) 8

Tardigrade Admin · Accepted Answer

P(x, y), Q(1,0) P Q=√(x-1)2+y2 P Q2=x2+1-2 x+y2 P Q2=x2-2 x+1+2 x3-3 x2+9 P Q2=2 x3-2 x2-2 x+10 (d P Q2/d x)=6 x2-4 x-2=0 6 x2-4 x-2=0 3 x2-2 x-1=0 3 x2-3 x+x-1=0 3 x(x-1)+(x-1)=0 (x-1)(3 x+1)=0 x=1, x=-(1/3) (d2 P Q2/d x2)=12 x-4 text So,PQ is minimum at x=1 y2=2+9-3 =8 (P Q) text minimum =√0+8=2 √2

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

$2$

$22$

$42$

$8$

Q. The minimum distance of the curve y2=2x3+9−3x2 from point Q(1,0) is

2

22​

42​

8

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

$2$

$22$

$42$

$8$