Consider the parabola P:y^2=4kx and the ellipse E:x^2a^2+y^2b^2=1.

Question

Consider the parabola $P:{y}^{2}=4kx$ 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 latus rectums. If the eccentricity of E is $e$, then ${e}^{2}+2\sqrt{2}$ is equal to $____$ .

Answer: 3

Accepted Answer

For the parabola

$P:{y}^{2}=4kx$

its latus rectum is the line

$x=k$

and the endpoints of the latus rectum are

$(k,\, 2k)\, \, (k,\, -2k).$

So, if the line segment joining the points of intersection of the parabola and ellipse is the latus rectum of the parabola, then the two curves intersect at

$(k,\pm 2k).$

Now for the ellipse

$E:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,$

its latus rectum is the chord through a focus perpendicular to the major axis.

Since the common chord is vertical, the major axis of the ellipse must be along the $x$-axis.

Hence the foci are

$(\pm ae,\, 0),$

where

$e=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}.$

The latus rectum of the ellipse is the vertical line through a focus:

$x=ae.$

Its endpoints are

$(ae,\pm \frac{{b}^{2}}{a}).$

But this same segment is also the latus rectum of the parabola, whose endpoints are

$(k,\pm 2k).$

Therefore, comparing coordinates:

$k=ae$

and

$2k=\frac{{b}^{2}}{a}.$

Substitute $k=ae$ into the second equation:

$2ae=\frac{{b}^{2}}{a}.$

So,

${b}^{2}=2{a}^{2}e.$

Now for the ellipse,

${b}^{2}={a}^{2}(1-{e}^{2}).$

Hence,

${a}^{2}(1-{e}^{2})=2{a}^{2}e.$

Dividing by ${a}^{2}$:

$1-{e}^{2}=2e.$

So,

${e}^{2}+2e-1=0.$

Since eccentricity is positive,

$e=-1+\sqrt{2}.$

Now calculate:

${e}^{2}+2\sqrt{2}.$

First,

${e}^{2}=(\sqrt{2}-1{)}^{2}=2+1-2\sqrt{2}=3-2\sqrt{2}.$

Therefore,

${e}^{2}+2\sqrt{2}=3.$

So the required value is

$3.$

Solutions · 1