Question

If one of the roots of the equation \(x^2-5x-14=0\) is the length of the semi-conjugate axis of the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] and the square of the other root is the semi-transverse axis, then the focus of the hyperbola that lies on the positive \(x\)-axis is

• A. \((5,0)\)
• B. \((\sqrt{65},0)\)
• C. \((7,0)\)
• D. \((\sqrt{74},0)\)

Hint:

For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the foci are \((\pm c,0)\), where \[ c^2=a^2+b^2. \]

Answer 1

The Correct Option is B

Step 1: Solve the quadratic using the formula.
\[x^2-5x-14=0 \implies x = \frac{5\pm\sqrt{25+56}}{2} = \frac{5\pm9}{2}\]
so the roots are \(x=7\) and \(x=-2\).

Step 2: Assign the roots to the axes.
A length has to be positive, so the semi-conjugate axis is the positive root, \(b=7\). The other root, \(-2\), squared gives the semi-transverse axis, \(a=(-2)^2=4\).

Step 3: Find c using \(c^2=a^2+b^2\).
\[c^2 = 4^2+7^2 = 16+49=65 \implies c=\sqrt{65}\]

Step 4: Locate the focus.
The transverse axis is along the x-axis, so the foci are at \((\pm c, 0)\), and the one on the positive x-axis is
\[\boxed{(\sqrt{65}, 0)}\]