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 $(\alpha, 2\sqrt{5})$ on the hyperbola is p, then 4p is equal to:

Step 1: Determine the center and orientation of the hyperbola

The given focus is $F_1 = (-5, 0)$ and the corresponding directrix is $x = -\frac{9}{5}$. Since the $y$ -coordinate of the focus is 0 and the directrix is a vertical line, the transverse axis of the hyperbola lies along the $x$ -axis. The center of the hyperbola is $(h, k) = (0, 0)$ because the focus is $(-c, 0)$ and the directrix is $x = -a/e$, which implies $c=5$ and $a/e = 9/5$.

Step 2: Calculate eccentricity and semi-transverse axis

For a hyperbola with center at the origin and transverse axis along the $x$ -axis, the focus is at $(-c, 0)$ and the directrix is $x = -a/e$. From the given information, we have $c = 5$ and $a/e = 9/5$. We can find the eccentricity $e$ by dividing $c$ by $a/e$, which gives $e = 5 / (9/5) = 25/9$. Then, we can find $a$ using $a = c/e = 5 / (25/9) = 9/5$.

Step 3: Calculate semi-conjugate axis

The relationship between $a$, $b$, and $e$ for a hyperbola is $b^2 = a^2(e^2 - 1)$. Substituting the values of $a = 9/5$ and $e = 25/9$, we calculate $b^2$. This gives $b^2 = (9/5)^2((25/9)^2 - 1) = (81/25)(625/81 - 1) = (81/25)(544/81) = 544/25$.

Step 4: Find the product of focal distances

For a point $P(x, y)$ on a hyperbola, the product of its focal distances is given by $PF_1 \cdot PF_2 = e^2x^2 - a^2$. We are given the point $P(\alpha, 2\sqrt{5})$. First, we substitute the coordinates of $P$ into the hyperbola's equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ to find $\alpha^2$. Then, we substitute $\alpha^2$, $e^2$, and $a^2$ into the focal distance product formula to find $p$.

Step 5: Calculate 4p

Finally, we need to calculate $4p$. We multiply the value of $p$ we found by 4.

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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