Let the product of the focal distances of the point P(4,2√3) on the hyperbola H:x^2a^2-y^2…

Question

Let the product of the focal distances of the point $P(4,2\sqrt{3})$ on the hyperbola $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ be 32 . Let the length of the conjugate axis of H be $p$ and the length of its latus rectum be $q$. Then ${p}^{2}+{q}^{2}$ is equal to\_\_\_\_\_\_\_\_\_\_

Answer: 120

Accepted Answer

To find ${p}^{2}+{q}^{2}$ for the hyperbola $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$, given that the product of the focal distances from a point $P(4,2\sqrt{3})$ to the foci is 32, we follow these steps:

Verify the Point on the Hyperbola:

The point $P(4,2\sqrt{3})$ satisfies the hyperbola equation:

$\frac{16}{{a}^{2}}-\frac{12}{{b}^{2}}=1\, \ldots$

Product of Focal Distances:

The distances from $P$ to the foci $S$ and ${S}^{'}$ are:

$SP=e(4-\frac{a}{e}),\, {S}^{'}P=e(4+\frac{a}{e})$

Therefore, the product:

$SP\cdot {S}^{'}P=16{e}^{2}-{a}^{2}=32$

Relating $e$, $a$, and $b$:

Using the identity for eccentricity:

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

Substituting in the product equation gives:

$16(1+\frac{{b}^{2}}{{a}^{2}})-{a}^{2}=32\, \ldots$

Solving for ${a}^{2}$ and ${b}^{2}$:

From equations (i) and (ii), solve for ${a}^{2}$ and ${b}^{2}$:

${a}^{2}=8,\, {b}^{2}=12$

Calculate ${p}^{2}+{q}^{2}$:

The conjugate axis length is $p=2b$ and the latus rectum is $q=\frac{2{b}^{2}}{a}$. Thus:

${p}^{2}+{q}^{2}=(2b{)}^{2}+{(\frac{2{b}^{2}}{a})}^{2}$

Substituting ${b}^{2}=12$ and ${a}^{2}=8$ gives:

${p}^{2}+{q}^{2}=4{b}^{2}+\frac{4{b}^{4}}{{a}^{2}}=48+\frac{4	imes 144}{8}$

${p}^{2}+{q}^{2}=48\cdot (1+\frac{12}{8})=120$

Therefore, ${p}^{2}+{q}^{2}=120$.

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