Question

find the equation of circle of curvature at (3, 6) on y^2=12x.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The equation of circle of curvature is $x^2+y^2-30x+12y-27=0$

Step-by-step explanation:

The equation of circle of curvature is

$(x-\overline x)^2+(y-\overline y)^2=R^2$

$\overline x=x-\dfrac{y_1}{y_2} \big(1+y_1^{2}\big )~~ \mbox{and}~~\overline y=y+\dfrac{1}{y_2} \big(1+y_1^{2}\big )$

where

$y_1=\dfrac{dy}{dx} ~~ \mbox{and}~~y_2=\dfrac{d^2y}{dx^2}$

And the radius of curvature R of a curve at a given point is the reciprocal of the curvature of the curve K at that point. i.e.,

$R=\dfrac{1}{K}$

And the radius of curvature is given by

$R=\dfrac{\Big(1+\big(y_1\big)^2\Big)^{3/2}}{\big| y_2\big|}$

Given the equation, $y^2=12x$

Differentiating with respect to $x$

$2y\dfrac{dy}{dx} =12$

$\implies \dfrac{dy}{dx} =\dfrac{12}{2y} =\dfrac{6}{y}$

$\implies \dfrac{d^2y}{dx^2} =-\dfrac{6}{y^2} \dfrac{dy}{dx}=-\dfrac{36}{y^3}$  

At point(3, 6), the respective values of the derivatives are given by

$y_1=\dfrac{dy}{dx}\bigg|_{3,6} =\dfrac{6}{6}=1$

$y_2=\dfrac{d^2y}{dx^2}\bigg|_{3,6} =-\dfrac{36}{y^3}=-\dfrac{36}{6^3}=-\dfrac{1}{6}$

Substituting the values, the radius of curvature at point(3, 6) is

$R=\dfrac{\Big(1+{1}^2\Big)^{3/2}}{ 1/{6}}={6\big(2\big)^{3/2}}=12\sqrt{2}$

And the values $\overline x ~~\mbox{and}~~\overline y$ at point(3, 6) are given by

$\overline x=x-\dfrac{y_1}{y_2} \big(1+y_1^{2}\big )=3-\dfrac{1}{-1/6} \big(1+1^{2}\big )=3+6\big(2\big )=15$

$\overline y=y+\dfrac{1}{y_2} \big(1+y_1^{2}\big )=6+\dfrac{1}{-1/6} \Big(1+1^{2}\Big )=6-6 \big(2\big)=-6$

Thus, the equation of circle of curvature is

$\big(x-15\big)^2+\big(y-(-6)\big)^2=\big(12\sqrt{2} \big)^2$

$\implies\big(x-15\big)^2+\big(y+6\big)^2=288$

$\implies x^2-30x+225+y^2+12y+36=288$

$\implies x^2+y^2-30x+12y-27=0$

Therefore, the equation of circle of curvature is $x^2+y^2-30x+12y-27=0$

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