Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2 …

Question

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $2 a$ and $2 b$, respectively, and one focus and the corresponding directrix of this hyperbola be $(-5,0)$ and $5 x+9=0$, respectively. If the product o...

Answer: 189

Accepted Answer

Given:

Transverse axis length: $2a$

Conjugate axis length: $2b$

One focus at $(-5,0)$

Directrix given by $5x+9=0$

The equations used are as follows:

Relationship between focus and directrix:

The focal length $ae=5$

The directrix gives $\frac{a}{e}=\frac{9}{5}$

Solving these equations, we get:

$ae=5,\, \frac{a}{e}=\frac{9}{5}$

From $ae=5$, we have $a=\frac{5}{e}$.

Substituting $a=3$ and $e=\frac{5}{3}$.

Finding $b$:

Use the relationship ${b}^{2}={a}^{2}({e}^{2}-1)$:

$a=3\, e=\frac{5}{3},\, b=4$

Equation of the hyperbola:

The standard equation, after substituting values of $a$ and $b$, is:

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

Focal distances product calculation:

For any point $(\alpha ,2\sqrt{5})$ that lies on the hyperbola:

$\frac{{\alpha }^{2}}{9}-\frac{(2\sqrt{5}{)}^{2}}{16}=1$

Solving this gives:

${\alpha }^{2}=9	imes \frac{36}{16}$

Product of focal distances ${PF}_{1}\cdot {PF}_{2}$:

${PF}_{1}\cdot {PF}_{2}=(e\alpha -a)(e\alpha +a)={e}^{2}{\alpha }^{2}-{a}^{2}$

Substituting the values:

$P={e}^{2}{\alpha }^{2}-{a}^{2}=\frac{25}{9}\cdot 9\cdot \frac{9}{4}-9=\frac{189}{4}$

Finally, to find $4p$:

$4p=4	imes \frac{189}{4}=189$

Solutions · 1