Let E:x^2a^2+y^2b^2=1,ab and H:x^2A^2-y^2B^2=1.

Question

Let $E:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>b$ and $H:\frac{{x}^{2}}{{A}^{2}}-\frac{{y}^{2}}{{B}^{2}}=1$. Let the distance between the foci of E and the foci of $H$ be $2\sqrt{3}$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$, then the sum of the lengths of their latus rectums is equal to :

Accepted Answer

We are given an ellipse

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

and a hyperbola

$\frac{{x}^{2}}{{A}^{2}}-\frac{{y}^{2}}{{B}^{2}}=1.$

It is stated that ``the distance between the foci of $E$ and the foci of $H$ is $2\sqrt{3}$.'' A natural interpretation is that the foci of the ellipse are separated by

$2\sqrt{{a}^{2}-{b}^{2}},$

and those of the hyperbola by

$2\sqrt{{A}^{2}+{B}^{2}},$

and each distance is equal to $2\sqrt{3}$. That is, we have

$2\sqrt{{a}^{2}-{b}^{2}}=2\sqrt{3}\, \implies \, {a}^{2}-{b}^{2}=3,$

and

$2\sqrt{{A}^{2}+{B}^{2}}=2\sqrt{3}\, \implies \, {A}^{2}+{B}^{2}=3.$

We are also given that

$a-A=2,$

and that the ratio of the eccentricities is

$\frac{{e}_{E}}{{e}_{H}}=\frac{1}{3},$

where the eccentricity of the ellipse is

${e}_{E}=\frac{\sqrt{{a}^{2}-{b}^{2}}}{a}=\frac{\sqrt{3}}{a},$

and the eccentricity of the hyperbola is

${e}_{H}=\frac{\sqrt{{A}^{2}+{B}^{2}}}{A}=\frac{\sqrt{3}}{A}.$

Thus the ratio becomes

$\frac{{e}_{E}}{{e}_{H}}=\frac{\sqrt{3}/a}{\sqrt{3}/A}=\frac{A}{a}=\frac{1}{3}.$

This implies

$a=3A.$

Now, using the condition $a-A=2$ together with $a=3A$, we get

$3A-A=2\, \implies \, 2A=2\, \implies \, A=1.$

Thus,

$a=3.$

Next, for the ellipse we have

${a}^{2}-{b}^{2}=3\, \implies \, 9-{b}^{2}=3\, \implies \, {b}^{2}=6.$

For the hyperbola,

${A}^{2}+{B}^{2}=3\, \implies \, 1+{B}^{2}=3\, \implies \, {B}^{2}=2.$

The length of the latus rectum is given by the following formulas:

For the ellipse:

${L}_{E}=\frac{2{b}^{2}}{a},$

For the hyperbola:

${L}_{H}=\frac{2{B}^{2}}{A}.$

Substitute the computed values:

For the ellipse:

${L}_{E}=\frac{2	imes 6}{3}=\frac{12}{3}=4,$

For the hyperbola:

${L}_{H}=\frac{2	imes 2}{1}=4.$

The sum of the lengths of the latus rectums is then

${L}_{E}+{L}_{H}=4+4=8.$

Thus, the answer is

$8.$

Solutions · 1