Let the eccentricity e of a hyperbola satisfy the equation 6e^2-11e+3=0.

Question

Let the eccentricity e of a hyperbola satisfy the equation $6{e}^{2}-11e+3=0$. If the foci of the hyperbola are $(3,5)$ and $(3,-4)$, then the length of its latus rectum is :

Accepted Answer

For a hyperbola, the eccentricity $e>1$.

We are given:

$6{e}^{2}-11e+3=0$

Solve this quadratic:

$6{e}^{2}-9e-2e+3=0$

$3e(2e-3)-1(2e-3)=0$

$(2e-3)(3e-1)=0$

So,

$e=\frac{3}{2}\, \, e=\frac{1}{3}$

Since for a hyperbola $e>1$, we take

$e=\frac{3}{2}$

Now the foci are $(3,5)$ and $(3,-4)$.

These lie on the vertical line $x=3$, so the transverse axis is vertical.

The distance between the foci is

$|5-(-4)|=9$

Hence,

$2c=9\Rightarrow c=\frac{9}{2}$

For a hyperbola,

$e=\frac{c}{a}$

So,

$a=\frac{c}{e}=\frac{\frac{9}{2}}{\frac{3}{2}}=3$

Now use the relation

${c}^{2}={a}^{2}+{b}^{2}$

Thus,

${b}^{2}={c}^{2}-{a}^{2}={(\frac{9}{2})}^{2}-{3}^{2}=\frac{81}{4}-9=\frac{81-36}{4}=\frac{45}{4}$

Length of latus rectum of a hyperbola is

$\frac{2{b}^{2}}{a}$

Therefore,

$=\frac{2\cdot \frac{45}{4}}{3}=\frac{45}{6}=\frac{15}{2}$

Hence, the correct answer is

$\frac{15}{2}$

So, Option C is correct.

Solutions · 1