Consider the circle C: x^2+y^2=4 and the para | vedclass

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(A) Let the point on the parabola $P: y^2=8x$ be $(x_1, y_1) = (2t^2, 4t)$.The equation of the chord of the circle $C: x^2+y^2=4$ bisected at $(x_1, y

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) Let the point on the parabola $P: y^2=8x$ be $(x_1, y_1) = (2t^2, 4t)$.The equation of the chord of the circle $C: x^2+y^2=4$ bisected at $(x_1, y_1)$ is given by $T=S_1$,where $T$ is $xx_1+yy_1-4$ and $S_1$ is $x_1^2+y_1^2-4$.So,$xx_1+yy_1 = x_1^2+y_1^2$.Since this chord passes through $(\alpha, 0)$,we have $\alpha x_1 = x_1^2+y_1^2$.Substituting $x_1=2t^2$ and $y_1=4t$,we get $\alpha(2t^2) = (2t^2)^2 + (4t)^2 = 4t^4 + 16t^2$.Dividing by $2t^2$ (assuming $t 
eq 0$),we get $\alpha = 2t^2 + 8$,which implies $t^2 = \frac{\alpha-8}{2}$.For the chord to exist within the circle,the midpoint $(x_1, y_1)$ must lie inside the circle,so $x_1^2+y_1^2 Substituting $x_1^2+y_1^2 = \alpha x_1 = \alpha(2t^2)$,we have $2\alpha t^2 Substituting $t^2 = \frac{\alpha-8}{2}$,we get $\alpha \left(\frac{\alpha-8}{2}ight) The roots of $\alpha^2 - 8\alpha - 4 = 0$ are $\alpha = \frac{8 \pm \sqrt{64+16}}{2} = 4 \pm 2\sqrt{5}$.Thus,$4-2\sqrt{5} Also,since $t^2 > 0$,we have $\frac{\alpha-8}{2} > 0$,so $\alpha > 8$.Combining these,$\alpha \in (8, 4+2\sqrt{5})$.Thus,$p=8$ and $q=4+2\sqrt{5}$.Then $(2q-p)^2 = (2(4+2\sqrt{5})-8)^2 = (8+4\sqrt{5}-8)^2 = (4\sqrt{5})^2 = 16 	imes 5 = 80$.

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