Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (−5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point (α, 2√5) on the hyperbola is p, then 4p is equal to:

 ❓ Question

Let a hyperbola in standard form have transverse and conjugate axes of lengths $2a$ and $2b$, respectively. One focus is at $(-5,0)$ and the corresponding directrix is $5x + 9 = 0$.

If the product of the focal distances of the point $(\alpha ,2\sqrt{5})\; on\; the\; hyperbola\; is$$p$, find $4p$.

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🖼️ Question Image

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✍️ Short Solution

1. Standard form of horizontal hyperbola:

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

Transverse axis = $2a$, conjugate axis = $2b$.

2. Use focus-directrix definition:

For a hyperbola, the eccentricity $e = c/a$ and distance from point $(x,y)$ to focus/focus-directrix property:

$$\text{Distance to focus} = e \cdot \text{distance to directrix}.$$

Given one focus $(-5,0)$ and directrix $5x+9=0 \implies x=-\frac{9}{5}$, eccentricity:

$$e=\frac{c}{a}\text{and }c=5\text{ (since focus at -5 and center at 0)}\mathrm{.}$$

Distance from a point $(x,y)$to the focus:

$$r_1=\sqrt{(x+5{)}^2+y^2},r_2=\sqrt{(x-5{)}^2+y^2}\mathrm{.}$$

3. Focal distance product:

Let the hyperbola be centered at origin. Then the product of distances of a point $(\alpha ,2\sqrt{5})\; to\; the\; two\; foci\; is\; known\; formula:$

$$(\text{distance to left focus})\cdot (\text{distance to right focus})=b^2+y^2$$

(This is a standard property of rectangular and general hyperbolas: for a point on $x$-axis symmetric hyperbola, product of focal distances = $b^2 + y^2$).

4. Compute $b^2$:

We know $c^2 = a^2 + b^2 \implies b^2 = c^2 - a^2 = 25 - a^2$.

Distance product:

$$p=b^2+y^2=(25-a^2)+(2\sqrt{5}{)}^2=(25-a^2)+20=45-a^2\mathrm{.}$$

5. Find $a$:

Focus at $(-5,0)$ and directrix $5x+9=0$ ⇒ distance from center to directrix $d = | -9/5| = 9/5$

Eccentricity $e=c\mathrm{/}a=5\mathrm{/}\mathrm{a\; = }$$c / |x_{\text{directrix}} - 0| = 5 / (9/5) = 25/9$

Check:

$$c=ae⟹a=c\mathrm{/}e=5\mathrm{/}(25\mathrm{/}9)=9\mathrm{/}5.$$

So $a^2=(9\mathrm{/}5{)}^2=81\mathrm{/}25$

6. Compute $p$:

$$p=45-a^2=45-\frac{81}{25}=\frac{1125-81}{25}=\frac{1044}{25}\mathrm{.}$$

7. Compute $4p$:

$$4p=4\cdot \frac{1044}{25}=\frac{4176}{25}\mathrm{.}$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

The product of focal distances for the point $(\alpha, 2\sqrt{5})$ is

$$p = \frac{1044}{25} \implies 4p = \frac{4176}{25}.$$
$$\boxed{4p = \frac{4176}{25}}$$