If (x, y) is a variable point on the curve x^ | vedclass

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(D) The given equation of the circle is $x^2 + y^2 - 2x - 2y - 2 = 0$.Completing the square,we get $(x - 1)^2 + (y - 1)^2 = 4$.Let $x - 1 = 2 \cos 	h

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The given equation of the circle is $x^2 + y^2 - 2x - 2y - 2 = 0$.Completing the square,we get $(x - 1)^2 + (y - 1)^2 = 4$.Let $x - 1 = 2 \cos 	heta$ and $y - 1 = 2 \sin 	heta$.Substituting these into the expression $E = \frac{8}{(x - 1)^2} - \frac{(y - 1)^2}{4}$,we get:$E = \frac{8}{4 \cos^2 	heta} - \frac{4 \sin^2 	heta}{4} = 2 \sec^2 	heta - \sin^2 	heta$.Using the identity $\sec^2 	heta = 1 + 	an^2 	heta$ and $\sin^2 	heta = 1 - \cos^2 	heta$,this does not simplify easily.Alternatively,since $(x - 1)^2 = 4 - (y - 1)^2$,let $u = (y - 1)^2$,where $0 \le u \le 4$.Then $E = \frac{8}{4 - u} - \frac{u}{4}$.Let $f(u) = \frac{8}{4 - u} - \frac{u}{4}$.$f'(u) = \frac{8}{(4 - u)^2} - \frac{1}{4}$.Setting $f'(u) = 0$,we get $(4 - u)^2 = 32$,so $4 - u = \pm 4 \sqrt{2}$.Since $0 \le u \le 4$,we check the boundaries.At $u = 0$,$E = \frac{8}{4} - 0 = 2$.At $u = 4$,the expression is undefined.As $u 	o 4^-$,$E 	o \infty$.Thus,the minimum value is $2$.

If $(x, y)$ is a variable point on the curve $x^2 + y^2 - 2x - 2y - 2 = 0$,then the minimum value of the expression $\frac{8}{(x - 1)^2} - \frac{(y - 1)^2}{4}$ is equal to

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