Consider the parabola P : y^2 = 4x and the el | vedclass

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(C) The parabola $P : y^2 = 4x$ has its focus at $(1, 0)$ and its latus rectum is the line $x = 1$.Since the line segment joining the points of inters

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$3$

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(C) The parabola $P : y^2 = 4x$ has its focus at $(1, 0)$ and its latus rectum is the line $x = 1$.Since the line segment joining the points of intersection of $P$ and $E$ is the latus rectum of both, the line $x = 1$ must be the latus rectum of the ellipse $E$.For the ellipse, the latus rectum is at $x = ae$, so $ae = 1$.The points of intersection are $(1, 2)$ and $(1, -2)$. Since these points lie on the ellipse, we substitute $x = 1$ and $y = 2$ into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:$\frac{1}{a^2} + \frac{4}{b^2} = 1$.Using the relation $b^2 = a^2(1 - e^2)$, we substitute $b^2$:$\frac{1}{a^2} + \frac{4}{a^2(1 - e^2)} = 1$.Since $a = 1/e$, we have $a^2 = 1/e^2$, so $e^2 + \frac{4e^2}{1 - e^2} = 1$.$e^2(1 - e^2) + 4e^2 = 1 - e^2 \implies e^2 - e^4 + 4e^2 = 1 - e^2 \implies e^4 - 6e^2 + 1 = 0$.Solving for $e^2$ using the quadratic formula: $e^2 = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2}$.Since $e Then $e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$.

Consider the parabola $P : y^2 = 4x$ and the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of $P$ and $E$ be their common latus rectum. If the eccentricity of $E$ is $e$, then $e^2 + 2\sqrt{2}$ is equal to . . . . . .

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