Let H:x^2a^2-y^2b^2=1 be a hyperbola such that the distance between its foci is 6 and the …

Question

Let $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is $\frac{8}{3}$. If the line $x=\alpha $ intersects the hyperbola H at the points A and B such that the area of the triangle AOB is $4\sqrt{15}$, where O is the origin, then ${\alpha }^{2}$ equals

Accepted Answer

The given equation of the hyperbola is:

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

Let the uniform eccentricity of the hyperbola be $e$.

According to the problem, the distance between the foci is $6$. For a standard hyperbola, the coordinates of the foci are $(\pm ae,0)$, so the distance between them is given by $2ae$.

$2ae=6$

$ae=3\, \ldots (i)$

The distance between the directrices is given as $\frac{8}{3}$. The equations of the directrices are $x=\pm \frac{a}{e}$, so the distance between them is $2\frac{a}{e}$.

$2\frac{a}{e}=\frac{8}{3}$

$\frac{a}{e}=\frac{4}{3}\, \ldots (ii)$

To find the value of ${a}^{2}$, we can multiply equations (i) and (ii):

$(ae)\cdot (\frac{a}{e})=3	imes \frac{4}{3}$

${a}^{2}=4$

We know the fundamental relationship between $a,b,$ and $e$ for a hyperbola is:

${b}^{2}={a}^{2}({e}^{2}-1)$

${b}^{2}={a}^{2}{e}^{2}-{a}^{2}$

Using the value $ae=3$, we have ${a}^{2}{e}^{2}=9$. Substituting ${a}^{2}{e}^{2}$ and ${a}^{2}$ into the equation gives:

${b}^{2}=9-4=5$

Now, substituting ${a}^{2}$ and ${b}^{2}$ back into the standard equation, the hyperbola is:

$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{5}=1$

The line $x=\alpha $ intersects the hyperbola at points A and B. Substitute $x=\alpha $ into the equation of the hyperbola to find the y-coordinates of points A and B:

$\frac{{\alpha }^{2}}{4}-\frac{{y}^{2}}{5}=1$

$\frac{{y}^{2}}{5}=\frac{{\alpha }^{2}}{4}-1$

$\frac{{y}^{2}}{5}=\frac{{\alpha }^{2}-4}{4}$

${y}^{2}=\frac{5({\alpha }^{2}-4)}{4}$

$y=\pm \frac{\sqrt{5({\alpha }^{2}-4)}}{2}$

So, the coordinates of points A and B are $(\alpha ,\frac{\sqrt{5({\alpha }^{2}-4)}}{2})$ and $(\alpha ,-\frac{\sqrt{5({\alpha }^{2}-4)}}{2})$.

These points lie vertically above and below each other, forming the base of $	riangleAOB$ along the line $x=\alpha $.

Length of the base $AB$ = Distance between A and B = $2	imes \frac{\sqrt{5({\alpha }^{2}-4)}}{2}=\sqrt{5({\alpha }^{2}-4)}$.

The perpendicular height of $	riangleAOB$ from the origin $O(0,0)$ to the line $x=\alpha $ is simply $|\alpha |$.

We are given that the area of $	riangleAOB$ is $4\sqrt{15}$. Using the area of a triangle formula:

$=\frac{1}{2}	imes 	imes$

$4\sqrt{15}=\frac{1}{2}\cdot \sqrt{5({\alpha }^{2}-4)}\cdot |\alpha |$

Squaring both sides to remove the square root and the absolute value:

$(4\sqrt{15}{)}^{2}=\frac{1}{4}\cdot ({\alpha }^{2})\cdot 5({\alpha }^{2}-4)$

$240=\frac{5}{4}{\alpha }^{2}({\alpha }^{2}-4)$

$240	imes \frac{4}{5}={\alpha }^{2}({\alpha }^{2}-4)$

$192={\alpha }^{4}-4{\alpha }^{2}$

${\alpha }^{4}-4{\alpha }^{2}-192=0$

Let ${\alpha }^{2}=t$. This transforms our equation into a quadratic equation:

${t}^{2}-4t-192=0$

Splitting the middle term:

${t}^{2}-16t+12t-192=0$

$t(t-16)+12(t-16)=0$

$(t-16)(t+12)=0$

So, $t=16$ or $t=-12$.

Since $t={\alpha }^{2}$ and the square of a real number cannot be negative, we disregard $-12$.

Thus, we have:

${\alpha }^{2}=16$

Option B is the correct answer.

Solutions · 1