Question

The foci of the hyperbola \( 16x^2 - 9y^2 - 64x + 18y - 90 = 0 \) are

• A. \( \left(\frac{24\pm5\sqrt{145}}{12},1\right) \)
• B. \( \left(\frac{21\pm5\sqrt{145}}{12},1\right) \)
• C. \( (1,\frac{24\pm5\sqrt{145}}{2}) \)
• D. \( (1,\frac{21\pm5\sqrt{145}}{2}) \)
• E. \( \left(\frac{21\pm5\sqrt{145}}{2},-1\right) \)

Hint:

Always complete squares first to convert conics into standard form.

Answer 1

The Correct Option is B

Concept: Standard hyperbola: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]

Step 1: Group terms.
\[ 16(x^2 -4x) -9(y^2 -2y) = 90 \]

Step 2: Complete square.
\[ 16[(x-2)^2 -4] -9[(y-1)^2 -1] = 90 \]

Step 3: Simplify.
\[ 16(x-2)^2 -9(y-1)^2 = 90 +64 -9 = 145 \]

Step 4: Standard form.
\[ \frac{(x-2)^2}{145/16} - \frac{(y-1)^2}{145/9} = 1 \]

Step 5: Compute \( c \).
\[ c^2 = a^2 + b^2 = \frac{145}{16} + \frac{145}{9} \Rightarrow c = \frac{5\sqrt{145}}{12} \] Focus: \[ \left(2 \pm \frac{5\sqrt{145}}{12}, 1\right) \]