Question

If the equation of the hyperbola having foci at \((8,3)\), \((0,3)\) and eccentricity \( \frac{4}{3} \) is \[ \frac{(x-\alpha)^2}{p} - \frac{(y-\beta)^2}{q} = 1, \] then find \( p + q \). Options:

• A. \( \beta^2 \)
• B. \( \alpha + \beta \)
• C. \( \alpha^2 \)
• D. \( \alpha \beta \)

Hint:

To solve hyperbola problems with given foci and eccentricity, find the center as midpoint of foci, calculate \(c\), use eccentricity to find \(a\), then find \(b\) using \(c^2 = a^2 + b^2\), and finally substitute into the standard equation.

Answer 1

The Correct Option is C

Step 1: Identify the center \((\alpha, \beta)\) of the hyperbola The foci are given as \((8,3)\) and \((0,3)\). Since the y-coordinates are the same, the center lies midway between the foci along the x-axis: \[ \alpha = \frac{8 + 0}{2} = 4, \quad \beta = \frac{3 + 3}{2} = 3. \] So, the center of the hyperbola is \((4,3)\). Step 2: Calculate the focal distance \(c\) \[ c = \text{distance from center to either focus} = 8 - 4 = 4. \] Step 3: Use eccentricity to find \(a\) \[ Eccentricity e = \frac{c}{a} = \frac{4}{3} \implies a = \frac{c}{e} = \frac{4}{\frac{4}{3}} = 3. \] Step 4: Find \(b\) using the hyperbola relation For hyperbola, \( c^2 = a^2 + b^2 \): \[ b^2 = c^2 - a^2 = 4^2 - 3^2 = 16 - 9 = 7. \] Step 5: Write the equation and identify \(p\) and \(q\) Since the transverse axis is horizontal (foci differ in x), the equation is: \[ \frac{(x-\alpha)^2}{a^2} - \frac{(y-\beta)^2}{b^2} = 1, \] thus: \[ p = a^2 = 9, \quad q = b^2 = 7. \] Step 6: Calculate \(p + q\) \[ p + q = 9 + 7 = 16. \] Note: \(\alpha = 4\), so \(\alpha^2 = 16\). Hence, \( p + q = \alpha^2 \).