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
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
Show Hint
For the hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\]
the foci are \((\pm c,0)\), where
\[
c^2=a^2+b^2.
\]
Updated On: Jul 18, 2026
- \((5,0)\)
- \((\sqrt{65},0)\)
- \((7,0)\)
- \((\sqrt{74},0)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Find the roots of the given quadratic equation.
Given equation is \[ x^2-5x-14=0 \] Factorizing, \[ x^2-7x+2x-14=0 \] \[ x(x-7)+2(x-7)=0 \] \[ (x+2)(x-7)=0 \] So, \[ x=-2 \quad \text{or} \quad x=7 \] The lengths of axes must be positive, so the positive root is \[ 7 \]
Step 2: Identify semi-conjugate and semi-transverse axes.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] the semi-transverse axis is \[ a \] and the semi-conjugate axis is \[ b \] According to the question, one root is the length of the semi-conjugate axis and the square of the other root is the semi-transverse axis.
Since the roots are \[ -2,\ 7 \] the length of the semi-conjugate axis must be \[ b=7 \] and the square of the other root gives \[ a=(-2)^2=4 \]
Step 3: Find the focus of the hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] we know that \[ c^2=a^2+b^2 \] Here, \[ a=4,\quad b=7 \] Therefore, \[ c^2=4^2+7^2 \] \[ c^2=16+49 \] \[ c^2=65 \] So, \[ c=\sqrt{65} \]
Step 4: Write the focus on the positive \(x\)-axis.
Since the transverse axis is along the \(x\)-axis, the foci are \[ (\pm c,0) \] Therefore, the focus lying on the positive \(x\)-axis is \[ (\sqrt{65},0) \]
Step 5: Final conclusion.
Therefore, \[ \boxed{(\sqrt{65},0)} \]
Given equation is \[ x^2-5x-14=0 \] Factorizing, \[ x^2-7x+2x-14=0 \] \[ x(x-7)+2(x-7)=0 \] \[ (x+2)(x-7)=0 \] So, \[ x=-2 \quad \text{or} \quad x=7 \] The lengths of axes must be positive, so the positive root is \[ 7 \]
Step 2: Identify semi-conjugate and semi-transverse axes.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] the semi-transverse axis is \[ a \] and the semi-conjugate axis is \[ b \] According to the question, one root is the length of the semi-conjugate axis and the square of the other root is the semi-transverse axis.
Since the roots are \[ -2,\ 7 \] the length of the semi-conjugate axis must be \[ b=7 \] and the square of the other root gives \[ a=(-2)^2=4 \]
Step 3: Find the focus of the hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] we know that \[ c^2=a^2+b^2 \] Here, \[ a=4,\quad b=7 \] Therefore, \[ c^2=4^2+7^2 \] \[ c^2=16+49 \] \[ c^2=65 \] So, \[ c=\sqrt{65} \]
Step 4: Write the focus on the positive \(x\)-axis.
Since the transverse axis is along the \(x\)-axis, the foci are \[ (\pm c,0) \] Therefore, the focus lying on the positive \(x\)-axis is \[ (\sqrt{65},0) \]
Step 5: Final conclusion.
Therefore, \[ \boxed{(\sqrt{65},0)} \]
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