Question:
Statement 1: The eccentricity of the hyperbola
\[
9x^2-16y^2-72x+96y-144=0
\]
is
\[
\frac{5}{4}.
\]
Statement 2: The eccentricity of the hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\]
is
\[
\sqrt{1+\frac{b^2}{a^2}}.
\]
Statement 1: The eccentricity of the hyperbola
\[
9x^2-16y^2-72x+96y-144=0
\]
is
\[
\frac{5}{4}.
\]
Statement 2: The eccentricity of the hyperbola
\[
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\]
is
\[
\sqrt{1+\frac{b^2}{a^2}}.
\]
Show Hint
For a hyperbola of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), use
\[
e=\sqrt{1+\frac{b^2}{a^2}}.
\]
Always convert the equation into standard form first.
Updated On: Jun 18, 2026
- Statement 1 is true, Statement 2 is true; Statement 2 is correct explanation for Statement 1.
- Both statements are true and Statement 2 is not the correct explanation of Statement 1.
- Statement 1 is false; Statement 2 is true.
- Statement 1 is true; Statement 2 is false.
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Convert the given hyperbola into standard form.
Given, \[ 9x^2-16y^2-72x+96y-144=0. \] Group the \(x\)-terms and \(y\)-terms: \[ 9(x^2-8x)-16(y^2-6y)-144=0. \] Completing the square, \[ x^2-8x=(x-4)^2-16 \] and \[ y^2-6y=(y-3)^2-9. \] Substitute these values: \[ 9[(x-4)^2-16]-16[(y-3)^2-9]-144=0. \] \[ 9(x-4)^2-144-16(y-3)^2+144-144=0. \] \[ 9(x-4)^2-16(y-3)^2=144. \] Dividing by \(144\), \[ \frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1. \]
Step 2: Identify \(a^2\) and \(b^2\).
Comparing with \[ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1, \] we get \[ a^2=16,\qquad b^2=9. \] Thus, \[ a=4,\qquad b=3. \]
Step 3: Use the eccentricity formula.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] eccentricity is \[ e=\sqrt{1+\frac{b^2}{a^2}}. \] So, \[ e=\sqrt{1+\frac{9}{16}}. \] \[ e=\sqrt{\frac{25}{16}}. \] \[ e=\frac{5}{4}. \]
Step 4: Verify both statements.
Statement 1 says that the eccentricity is \[ \frac{5}{4}, \] which is true.
Statement 2 gives the correct formula for the eccentricity of the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] which is also true.
Also, Statement 2 is the correct explanation for Statement 1.
Step 5: Final conclusion.
Therefore, \[ \boxed{\text{Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation for Statement 1.}} \]
Given, \[ 9x^2-16y^2-72x+96y-144=0. \] Group the \(x\)-terms and \(y\)-terms: \[ 9(x^2-8x)-16(y^2-6y)-144=0. \] Completing the square, \[ x^2-8x=(x-4)^2-16 \] and \[ y^2-6y=(y-3)^2-9. \] Substitute these values: \[ 9[(x-4)^2-16]-16[(y-3)^2-9]-144=0. \] \[ 9(x-4)^2-144-16(y-3)^2+144-144=0. \] \[ 9(x-4)^2-16(y-3)^2=144. \] Dividing by \(144\), \[ \frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1. \]
Step 2: Identify \(a^2\) and \(b^2\).
Comparing with \[ \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1, \] we get \[ a^2=16,\qquad b^2=9. \] Thus, \[ a=4,\qquad b=3. \]
Step 3: Use the eccentricity formula.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] eccentricity is \[ e=\sqrt{1+\frac{b^2}{a^2}}. \] So, \[ e=\sqrt{1+\frac{9}{16}}. \] \[ e=\sqrt{\frac{25}{16}}. \] \[ e=\frac{5}{4}. \]
Step 4: Verify both statements.
Statement 1 says that the eccentricity is \[ \frac{5}{4}, \] which is true.
Statement 2 gives the correct formula for the eccentricity of the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] which is also true.
Also, Statement 2 is the correct explanation for Statement 1.
Step 5: Final conclusion.
Therefore, \[ \boxed{\text{Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation for Statement 1.}} \]
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