Question:
If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola
\[
16x^2-9y^2=1
\]
and its conjugate respectively, then \(3e_1=\)
If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola
\[
16x^2-9y^2=1
\]
and its conjugate respectively, then \(3e_1=\)
Show Hint
For a hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,
\]
the eccentricity is
\[
e=\sqrt{1+\frac{b^2}{a^2}}
\]
Always rewrite the equation in standard form before applying formulas.
Updated On: Jun 22, 2026
- \(5e_2\)
- \(4e_2\)
- \(2e_2\)
- \(e_2\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write the hyperbola in standard form.
Given, \[ 16x^2-9y^2=1 \] Divide throughout by \(1\), \[ \frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1 \] Comparing with \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] we get \[ a^2=\frac{1}{16}, \qquad b^2=\frac{1}{9} \]
Step 2: Find the eccentricity \(e_1\) of the hyperbola.
For a hyperbola, \[ e_1=\sqrt{1+\frac{b^2}{a^2}} \] Substituting the values, \[ e_1=\sqrt{1+\frac{\frac{1}{9}}{\frac{1}{16}}} \] \[ e_1=\sqrt{1+\frac{16}{9}} \] \[ e_1=\sqrt{\frac{25}{9}} \] \[ e_1=\frac{5}{3} \]
Step 3: Find the eccentricity \(e_2\) of the conjugate hyperbola.
The conjugate hyperbola is \[ \frac{y^2}{\frac{1}{9}}-\frac{x^2}{\frac{1}{16}}=1 \] Hence, \[ a^2=\frac{1}{9}, \qquad b^2=\frac{1}{16} \] Its eccentricity is \[ e_2=\sqrt{1+\frac{b^2}{a^2}} \] So, \[ e_2=\sqrt{1+\frac{\frac{1}{16}}{\frac{1}{9}}} \] \[ e_2=\sqrt{1+\frac{9}{16}} \] \[ e_2=\sqrt{\frac{25}{16}} \] \[ e_2=\frac{5}{4} \]
Step 4: Compare \(3e_1\) and \(e_2\).
Now, \[ 3e_1=3\times \frac{5}{3}=5 \] Also, \[ 4e_2=4\times \frac{5}{4}=5 \] Thus, \[ 3e_1=4e_2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{3e_1=4e_2} \]
Given, \[ 16x^2-9y^2=1 \] Divide throughout by \(1\), \[ \frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1 \] Comparing with \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] we get \[ a^2=\frac{1}{16}, \qquad b^2=\frac{1}{9} \]
Step 2: Find the eccentricity \(e_1\) of the hyperbola.
For a hyperbola, \[ e_1=\sqrt{1+\frac{b^2}{a^2}} \] Substituting the values, \[ e_1=\sqrt{1+\frac{\frac{1}{9}}{\frac{1}{16}}} \] \[ e_1=\sqrt{1+\frac{16}{9}} \] \[ e_1=\sqrt{\frac{25}{9}} \] \[ e_1=\frac{5}{3} \]
Step 3: Find the eccentricity \(e_2\) of the conjugate hyperbola.
The conjugate hyperbola is \[ \frac{y^2}{\frac{1}{9}}-\frac{x^2}{\frac{1}{16}}=1 \] Hence, \[ a^2=\frac{1}{9}, \qquad b^2=\frac{1}{16} \] Its eccentricity is \[ e_2=\sqrt{1+\frac{b^2}{a^2}} \] So, \[ e_2=\sqrt{1+\frac{\frac{1}{16}}{\frac{1}{9}}} \] \[ e_2=\sqrt{1+\frac{9}{16}} \] \[ e_2=\sqrt{\frac{25}{16}} \] \[ e_2=\frac{5}{4} \]
Step 4: Compare \(3e_1\) and \(e_2\).
Now, \[ 3e_1=3\times \frac{5}{3}=5 \] Also, \[ 4e_2=4\times \frac{5}{4}=5 \] Thus, \[ 3e_1=4e_2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{3e_1=4e_2} \]
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