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

Question

If the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, passing through $(6,4 \sqrt{3})$, satisfies $15\left(e^2+1\right)=34 e$

Accepted Answer

For the hyperbola

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$

we are given that it passes through the point $(6,4\sqrt{3})$.

So, substituting $x=6$ and $y=4\sqrt{3}$:

$\frac{36}{{a}^{2}}-\frac{(4\sqrt{3}{)}^{2}}{{b}^{2}}=1$

Since

$(4\sqrt{3}{)}^{2}=48$

we get

$\frac{36}{{a}^{2}}-\frac{48}{{b}^{2}}=1\, ...(1)$

Now for the hyperbola

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$

its eccentricity is

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

So,

${e}^{2}=1+\frac{{b}^{2}}{{a}^{2}}$

Given:

$15({e}^{2}+1)=34e$

Substitute ${e}^{2}=1+\frac{{b}^{2}}{{a}^{2}}$ if needed, but first solve this equation in $e$:

$15{e}^{2}-34e+15=0$

Factorising:

$15{e}^{2}-25e-9e+15=0$

$5e(3e-5)-3(3e-5)=0$

$(3e-5)(5e-3)=0$

So,

$e=\frac{5}{3}\, \, e=\frac{3}{5}$

But for a hyperbola, $e>1$, hence

$e=\frac{5}{3}$

Now,

${e}^{2}=1+\frac{{b}^{2}}{{a}^{2}}$

So,

$\frac{25}{9}=1+\frac{{b}^{2}}{{a}^{2}}$

$\frac{{b}^{2}}{{a}^{2}}=\frac{25}{9}-1=\frac{16}{9}$

Thus,

${b}^{2}=\frac{16}{9}{a}^{2}$

Substitute this in equation $(1)$:

$\frac{36}{{a}^{2}}-\frac{48}{\frac{16}{9}{a}^{2}}=1$

Now,

$\frac{48}{\frac{16}{9}{a}^{2}}=48\cdot \frac{9}{16{a}^{2}}=\frac{27}{{a}^{2}}$

So,

$\frac{36}{{a}^{2}}-\frac{27}{{a}^{2}}=1$

$\frac{9}{{a}^{2}}=1$

Hence,

${a}^{2}=9$

Then,

${b}^{2}=\frac{16}{9}\cdot 9=16$

Now consider the second hyperbola:

$\frac{{x}^{2}}{{b}^{2}}-\frac{{y}^{2}}{2({a}^{2}+1)}=1$

Substitute ${a}^{2}=9$ and ${b}^{2}=16$:

$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{2(9+1)}=1$

$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{20}=1$

For a hyperbola of the form

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

the length of the latus rectum is

$\frac{2{B}^{2}}{A}$

Here,

${A}^{2}=16\Rightarrow A=4,\, {B}^{2}=20$

Therefore, length of latus rectum

$=\frac{2\cdot 20}{4}=10$

Hence, the required answer is

$10$

So, the correct option is A.

Solutions · 1