Question:
The eccentricity of the hyperbola
\[
16x^2-9y^2=144
\]
is
The eccentricity of the hyperbola
\[
16x^2-9y^2=144
\]
is
Show Hint
For a hyperbola,
\[
e=\frac{c}{a}
\]
and
\[
c^2=a^2+b^2.
\]
Combining these gives
\[
e=\sqrt{1+\frac{b^2}{a^2}}.
\]
Updated On: Jun 10, 2026
- \(\dfrac{5}{4}\)
- \(\dfrac{3}{2}\)
- \(\dfrac{4}{3}\)
- \(\dfrac{5}{3}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The standard form of a hyperbola is
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.
\]
Its eccentricity is
\[
e=\sqrt{1+\frac{b^2}{a^2}}.
\]
Step 1: Convert into standard form Given \[ 16x^2-9y^2=144. \] Dividing throughout by \(144\), \[ \frac{x^2}{9}-\frac{y^2}{16}=1. \] Thus, \[ a^2=9, \qquad b^2=16. \]
Step 2: Calculate eccentricity \[ e = \sqrt{1+\frac{16}{9}}. \] \[ = \sqrt{\frac{25}{9}}. \] \[ = \frac53. \] Hence, \[ \boxed{\frac53}. \]
Step 3: Verify with the options The mathematically correct value is \[ \boxed{\frac53}. \] Thus the correct option is \[ \boxed{(D)}. \]
Step 1: Convert into standard form Given \[ 16x^2-9y^2=144. \] Dividing throughout by \(144\), \[ \frac{x^2}{9}-\frac{y^2}{16}=1. \] Thus, \[ a^2=9, \qquad b^2=16. \]
Step 2: Calculate eccentricity \[ e = \sqrt{1+\frac{16}{9}}. \] \[ = \sqrt{\frac{25}{9}}. \] \[ = \frac53. \] Hence, \[ \boxed{\frac53}. \]
Step 3: Verify with the options The mathematically correct value is \[ \boxed{\frac53}. \] Thus the correct option is \[ \boxed{(D)}. \]
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