Let the foci of hyperbola coincide with the foci of the ellipse fracx^236+fracy^216=1. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is:

Let e_1 be eccentricity of ellipse \ => e_1=sqrt(1-(16 / 36))=sqrt(1-(4 / 9))=fracsqrt(5)3 So ae_1=6 * fracsqrt(5)3=2 sqrt(5) Now H : fracx^2p^2-fracy^2q^2=1 p.e = ae_1 p. 5=2 sqrt(5) p=frac2sqrt(5) => e^2 =1+frac q^2 p^2 => 25=1+frac5 q^24 \ => q^2 =(96 / 5) So length of L R frac2 q^2p=frac96sqrt(5)

Let the foci of hyperbola coincide with the foci of the... | null (Mathematics)

Let

\mathrm{e}_{1}

be eccentricity of ellipse

\\ \Rightarrow e_{1}=\sqrt{1-\frac{16}{36}}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\\

\mathrm{ae}_{1}=6 \cdot \frac{\sqrt{5}}{3}=2 \sqrt{5}\\

Now H :

\frac{x^{2}}{p^{2}}-\frac{y^{2}}{q^{2}}=1\\

p.e

=\mathrm{ae}_{1}\\

5=2 \sqrt{5}\\

\mathrm{p}=\frac{2}{\sqrt{5}} \Rightarrow \mathrm{e}^{2} \\=1+\frac{\mathrm{q}^{2}}{\mathrm{p}^{2}} \Rightarrow 25=1+\frac{5 \mathrm{q}^{2}}{4} \\ \Rightarrow \mathrm{q}^{2} =\frac{96}{5}\\

So length of

L R \frac{2 q^{2}}{p}=\frac{96}{\sqrt{5}}

Explanation

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