Question:
Let
\[
x^2+y^2=16
\]
be the equation of the auxiliary circle of a hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\]
and let \((4\sqrt{2},3)\) be a point on the hyperbola. Then the eccentricity of the hyperbola is
Let
\[
x^2+y^2=16
\]
be the equation of the auxiliary circle of a hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\]
and let \((4\sqrt{2},3)\) be a point on the hyperbola. Then the eccentricity of the hyperbola is
Show Hint
For the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), the auxiliary circle is \(x^2+y^2=a^2\), and eccentricity is \(e=\sqrt{1+\frac{b^2}{a^2}}\).
Updated On: Jun 26, 2026
- \(\frac{5}{4}\)
- \(\frac{5}{3}\)
- \(\frac{4}{3}\)
- \(2\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use the auxiliary circle of the hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the auxiliary circle is \[ x^2+y^2=a^2 \] Given auxiliary circle is \[ x^2+y^2=16 \] Therefore, \[ a^2=16 \] So, \[ a=4 \]
Step 2: Use the point lying on the hyperbola.
The point \[ (4\sqrt{2},3) \] lies on the hyperbola. Substitute \[ x=4\sqrt{2},\quad y=3,\quad a^2=16 \] in \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] We get \[ \frac{(4\sqrt{2})^2}{16}-\frac{3^2}{b^2}=1 \] \[ \frac{32}{16}-\frac{9}{b^2}=1 \] \[ 2-\frac{9}{b^2}=1 \] \[ \frac{9}{b^2}=1 \] Hence, \[ b^2=9 \]
Step 3: Find the eccentricity of the hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] eccentricity is given by \[ e=\sqrt{1+\frac{b^2}{a^2}} \] Substituting \[ a^2=16,\quad b^2=9 \] we get \[ e=\sqrt{1+\frac{9}{16}} \] \[ e=\sqrt{\frac{25}{16}} \] \[ e=\frac{5}{4} \]
Step 4: Final conclusion.
Therefore, the eccentricity of the hyperbola is \[ \boxed{\frac{5}{4}} \]
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the auxiliary circle is \[ x^2+y^2=a^2 \] Given auxiliary circle is \[ x^2+y^2=16 \] Therefore, \[ a^2=16 \] So, \[ a=4 \]
Step 2: Use the point lying on the hyperbola.
The point \[ (4\sqrt{2},3) \] lies on the hyperbola. Substitute \[ x=4\sqrt{2},\quad y=3,\quad a^2=16 \] in \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] We get \[ \frac{(4\sqrt{2})^2}{16}-\frac{3^2}{b^2}=1 \] \[ \frac{32}{16}-\frac{9}{b^2}=1 \] \[ 2-\frac{9}{b^2}=1 \] \[ \frac{9}{b^2}=1 \] Hence, \[ b^2=9 \]
Step 3: Find the eccentricity of the hyperbola.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] eccentricity is given by \[ e=\sqrt{1+\frac{b^2}{a^2}} \] Substituting \[ a^2=16,\quad b^2=9 \] we get \[ e=\sqrt{1+\frac{9}{16}} \] \[ e=\sqrt{\frac{25}{16}} \] \[ e=\frac{5}{4} \]
Step 4: Final conclusion.
Therefore, the eccentricity of the hyperbola is \[ \boxed{\frac{5}{4}} \]
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