If the eccentricity of the hyperbola x^2a^2-y^2b^2=1, passing through (6,4√3), satisfies 1…

Question

If the eccentricity of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, passing through $(6,4\sqrt3)$, satisfies $15(e^2+1)=34e$, then the length of the latus rectum of the hyperbola $\dfrac{x^2}{b^2}-\dfrac{y^2}{2(a^2+1)}=1$ is:

Accepted Answer

$15e^2-34e+15=0 \Rightarrow (3e-5)(5e-3)=0 \Rightarrow e=\dfrac53$ (taking $e>1$).

$$e^2=\frac{25}{9}=1+\frac{b^2}{a^2}\;\Rightarrow\;b^2=\frac{16}{9}a^2.$$

Through $(6,4\sqrt3)$: $\dfrac{36}{a^2}-\dfrac{48}{b^2}=1 \Rightarrow \dfrac{36}{a^2}-\dfrac{27}{a^2}=1 \Rightarrow a^2=9,\ b^2=16$.

Second hyperbola: $\dfrac{x^2}{16}-\dfrac{y^2}{2(10)}=1=\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$, so $A^2=16,\ B^2=20$.

$$\text{Latus rectum}=\frac{2B^2}{A}=\frac{2\times20}{4}=10.$$

Solutions · 1