Let P be the parabola in the plane determined | vedclass

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Question

(B) The equation of the parabola is $y = x^2$. $A$ circle $x^2 + y^2 + Dx + Ey + F = 0$ intersects the parabola at points where $y + y^2 + Dx + Ey + F

Vedclass · Accepted Answer

$1247$

Vedclass · Answer

(B) The equation of the parabola is $y = x^2$. $A$ circle $x^2 + y^2 + Dx + Ey + F = 0$ intersects the parabola at points where $y + y^2 + Dx + Ey + F = 0$. Substituting $x^2 = y$,we get $y^2 + (1+E)y + Dx + F = 0$. Since $x = \pm \sqrt{y}$,the equation becomes $y^2 + (1+E)y + F = \pm D\sqrt{y}$,which leads to $(y^2 + (1+E)y + F)^2 = D^2y$. This is a quartic in $y$,so the sum of the $y$-coordinates of the four intersection points is $y_1 + y_2 + y_3 + y_4 = -(1+E)^2 + 2F$. However,a simpler property for a circle intersecting $y=x^2$ is that the sum of the $x$-coordinates of the four points is $0$. Let the points be $(x_1, x_1^2), (x_2, x_2^2), (x_3, x_3^2), (x_4, x_4^2)$. Given $x_1 = 17, x_2 = -2, x_3 = 13$,we have $17 - 2 + 13 + x_4 = 0$,which gives $x_4 = -28$. The fourth point is $(-28, 784)$. The directrix of $y = x^2$ is $y = -\frac{1}{4}$. The perpendicular distance from a point $(x_i, y_i)$ to the line $y = -\frac{1}{4}$ is $y_i - (-\frac{1}{4}) = y_i + \frac{1}{4}$. The sum of the distances is $\sum_{i=1}^{4} (y_i + \frac{1}{4}) = (y_1 + y_2 + y_3 + y_4) + 4(\frac{1}{4}) = (289 + 4 + 169 + 784) + 1 = 1246 + 1 = 1247$.

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