Question:
The foci of the ellipse
\[
9x^2+25y^2=225
\]
are
The foci of the ellipse
\[
9x^2+25y^2=225
\]
are
Show Hint
For the ellipse
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
\]
the foci are obtained using
\[
c^2=a^2-b^2.
\]
If the larger denominator is under \(x^2\), then the foci are \((\pm c,0)\).
Updated On: Jun 26, 2026
- \((\pm 4,0)\)
- \(\left(\pm \frac{4}{5},0\right)\)
- \(\left(\pm \frac{12}{5},0\right)\)
- \(\left(\pm \frac{2}{5},0\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Convert the ellipse into standard form.
Given: \[ 9x^2+25y^2=225 \] Dividing throughout by \(225\), \[ \frac{x^2}{25}+\frac{y^2}{9}=1 \] Comparing with the standard form \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] we obtain \[ a^2=25,\qquad b^2=9. \] Hence, \[ a=5,\qquad b=3. \]
Step 2: Find the value of \(c\).
For an ellipse, \[ c^2=a^2-b^2. \] Therefore, \[ c^2=25-9=16. \] Hence, \[ c=4. \]
Step 3: Determine the foci.
Since \[ a^2\gt b^2, \] the major axis lies along the \(x\)-axis.
Therefore, the foci are \[ (\pm c,0). \] Substituting \(c=4\), \[ (\pm 4,0). \]
Step 4: Final conclusion.
Hence, the foci of the ellipse are \[ \boxed{(\pm 4,0)}. \]
Given: \[ 9x^2+25y^2=225 \] Dividing throughout by \(225\), \[ \frac{x^2}{25}+\frac{y^2}{9}=1 \] Comparing with the standard form \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] we obtain \[ a^2=25,\qquad b^2=9. \] Hence, \[ a=5,\qquad b=3. \]
Step 2: Find the value of \(c\).
For an ellipse, \[ c^2=a^2-b^2. \] Therefore, \[ c^2=25-9=16. \] Hence, \[ c=4. \]
Step 3: Determine the foci.
Since \[ a^2\gt b^2, \] the major axis lies along the \(x\)-axis.
Therefore, the foci are \[ (\pm c,0). \] Substituting \(c=4\), \[ (\pm 4,0). \]
Step 4: Final conclusion.
Hence, the foci of the ellipse are \[ \boxed{(\pm 4,0)}. \]
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