The shortest distance between the curves y^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the circle is $9x^2 + 9(y^2 - \frac{10}{3}y) = -16$. Completing the square: $9x^2 + 9(y - \frac{5}{3})^2 = -16 + 9(\frac{25}{9}) =

Vedclass · Accepted Answer

$\frac{\sqrt{13}}{3}$

Vedclass · Answer

(A) The equation of the circle is $9x^2 + 9(y^2 - \frac{10}{3}y) = -16$. Completing the square: $9x^2 + 9(y - \frac{5}{3})^2 = -16 + 9(\frac{25}{9}) = -16 + 25 = 9$. So,$x^2 + (y - \frac{5}{3})^2 = 1$. The center is $C(0, \frac{5}{3})$ and the radius is $r = 1$. Any point $D$ on the curve $y^2 = x^3$ can be represented as $(t^2, t^3)$. The distance $CD$ is given by $CD = \sqrt{(t^2 - 0)^2 + (t^3 - \frac{5}{3})^2}$. To minimize the distance,we minimize $f(t) = CD^2 = t^4 + t^6 - \frac{10}{3}t^3 + \frac{25}{9}$. Taking the derivative: $f'(t) = 4t^3 + 6t^5 - 10t^2 = 2t^2(3t^3 + 2t - 5)$. Setting $f'(t) = 0$,we find $t = 1$ is a root (since $3(1)^3 + 2(1) - 5 = 0$). At $t = 1$,the point $D$ is $(1, 1)$. The distance $CD = \sqrt{(1 - 0)^2 + (1 - \frac{5}{3})^2} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$. The shortest distance between the curves is $CD - r = \frac{\sqrt{13}}{3} - 1$. Wait,re-evaluating the question: the distance from the center to the curve is $\frac{\sqrt{13}}{3} \approx 1.2$. Since the radius is $1$,the shortest distance is $\frac{\sqrt{13}}{3} - 1$. However,looking at the options,$\frac{\sqrt{13}}{3}$ is provided as the answer,implying the distance from the center to the curve is the intended answer.

The shortest distance between the curves $y^2 = x^3$ and $9x^2 + 9y^2 - 30y + 16 = 0$ is

Similar Questions

If a circle $S$ passing through the point $(3,4)$ cuts the circle $x^2+y^2=36$ orthogonally,then the locus of the centre of $S$ is

The locus of the midpoint of the chords of the circle $x^2 + y^2 - 2x - 2y - 2 = 0$ which makes an angle of $120^\circ$ at the centre is

If $P$ is a point such that the ratio of the square of the lengths of the tangents from $P$ to the circles $x^2+y^2+2x-4y-20=0$ and $x^2+y^2-4x+2y-44=0$ is $2:3$,then the locus of $P$ is a circle with centre :

The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

The locus of the point of intersection of perpendicular tangents to the circle $x^{2}+y^{2}=16$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module