Question

If the foci of the ellipse \( \frac{x^2}{16}+\frac{y^2}{b^2}=1 \) and the foci of the hyperbola \( \frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} \) coincide, then the value of \( b^2 \) is

• A. 5
• B. 9
• C. 1
• D. 7

Hint:

For ellipse, use \( c^2=a^2-b^2 \), while for hyperbola, use \( c^2=a^2+b^2 \)Always convert equation to standard form first.

Answer 1

The Correct Option is D

Step 1: Write the given ellipse.
\[ \frac{x^2}{16}+\frac{y^2}{b^2}=1 \]
For this ellipse:
\[ a^2=16 \]

Step 2: Write the given hyperbola in standard form.
\[ \frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} \]
Multiply both sides by \(25\):
\[ \frac{25x^2}{144}-\frac{25y^2}{81}=1 \]
\[ \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1 \]

Step 3: Identify values for hyperbola.
For hyperbola:
\[ a^2=\frac{144}{25},\quad b^2=\frac{81}{25} \]

Step 4: Find \(c^2\) for hyperbola.
For hyperbola:
\[ c^2=a^2+b^2 \]
\[ c^2=\frac{144}{25}+\frac{81}{25} \]
\[ c^2=\frac{225}{25}=9 \]

Step 5: Use coinciding foci condition.
Since the foci of ellipse and hyperbola coincide, ellipse must have same \(c^2\).
So for ellipse:
\[ c^2=9 \]

Step 6: Apply ellipse formula.
For ellipse:
\[ c^2=a^2-b^2 \]
\[ 9=16-b^2 \]

Step 7: Final conclusion.
\[ b^2=16-9=7 \]
\[ \boxed{7} \]