Let $\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ and $\mathrm{H}: \frac{x^2}{\mathrm{~A}^2}-\frac{y^2}{\mathrm{~B}^2}=1$. 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 :

<p>We are given an ellipse</p> <p>$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b,$</p> <p>and a hyperbola</p> <p>$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.$</p> <p>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 </p> <p>$2\sqrt{a^2-b^2},$ </p> <p>and those of the hyperbola by </p> <p>$2\sqrt{A^2+B^2},$ </p> <p>and each distance is equal to $2\sqrt{3}$. That is, we have</p> <p>$ 2\sqrt{a^2-b^2}=2\sqrt{3}\quad \Longrightarrow\quad a^2-b^2=3, $</p> <p>and</p> <p>$ 2\sqrt{A^2+B^2}=2\sqrt{3}\quad \Longrightarrow\quad A^2+B^2=3. $</p> <p>We are also given that</p> <p>$a-A=2,$</p> <p>and that the ratio of the eccentricities is</p> <p>$\frac{e_E}{e_H}=\frac{1}{3},$</p> <p>where the eccentricity of the ellipse is </p> <p>$e_E=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{3}}{a},$</p> <p>and the eccentricity of the hyperbola is </p> <p>$e_H=\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{3}}{A}.$</p> <p>Thus the ratio becomes</p> <p>$\frac{e_E}{e_H}=\frac{\sqrt{3}/a}{\sqrt{3}/A}=\frac{A}{a}=\frac{1}{3}.$</p> <p>This implies</p> <p>$a=3A.$</p> <p>Now, using the condition $a-A=2$ together with $a=3A$, we get</p> <p>$3A-A=2 \quad\Longrightarrow\quad 2A=2 \quad\Longrightarrow\quad A=1.$</p> <p>Thus,</p> <p>$a=3.$</p> <p>Next, for the ellipse we have</p> <p>$a^2-b^2=3 \quad \Longrightarrow\quad 9-b^2=3 \quad\Longrightarrow\quad b^2=6.$</p> <p>For the hyperbola,</p> <p>$A^2+B^2=3 \quad \Longrightarrow\quad 1+B^2=3 \quad\Longrightarrow\quad B^2=2.$</p> <p>The length of the latus rectum is given by the following formulas:</p> <p>For the ellipse:</p> <p>$L_E=\frac{2b^2}{a},$</p> <p>For the hyperbola:</p> <p>$L_H=\frac{2B^2}{A}.$</p> <p>Substitute the computed values:</p> <p>For the ellipse:</p> <p>$L_E=\frac{2\times6}{3}=\frac{12}{3}=4,$</p> <p>For the hyperbola:</p> <p>$L_H=\frac{2\times2}{1}=4.$</p> <p>The sum of the lengths of the latus rectums is then</p> <p>$L_E+L_H=4+4=8.$</p> <p>Thus, the answer is</p> <p>$8.$</p>