Question

If \(d_1\) and \(d_2\) are the distances of the foci of the hyperbola \(4x^{2}-9y^{2}-16x+54y-101=0\) from the point (2,-3), then \(d_1+d_2=\)

• A. 10
• B. 14
• C. 12
• D. 16

Hint:

When the point given for distance calculation lies on the minor axis (in this case, the line \(x=h=2\)), the calculation of distances to the foci simplifies significantly because the horizontal offsets cancel out.

Answer 1

The Correct Option is B

Step 1: Goal and given.
We have the hyperbola $4x^2-9y^2-16x+54y-101=0$ and a point $(2,-3)$. We want $d_1+d_2$, the sum of distances from this point to the two foci.
Step 2: Complete the squares.
Group: $4(x^2-4x)-9(y^2-6y)=101$. Completing, \[ 4(x-2)^2-9(y-3)^2=101+16-81=36. \]
Step 3: Get standard form.
Dividing by $36$: $\dfrac{(x-2)^2}{9}-\dfrac{(y-3)^2}{4}=1$. So $A^2=9,\,B^2=4$, centre $(2,3)$.
Step 4: Find the foci.
Here $A=3$ and $e=\sqrt{1+\tfrac{B^2}{A^2}}=\sqrt{1+\tfrac49}=\dfrac{\sqrt{13}}{3}$, so $Ae=\sqrt{13}$. The foci are $(2\pm\sqrt{13},\,3)$.
Step 5: Distance from $(2,-3)$ to each focus.
The horizontal gap is $\pm\sqrt{13}$ and the vertical gap is $3-(-3)=6$ for both. So \[ d_1=d_2=\sqrt{(\sqrt{13})^2+6^2}=\sqrt{13+36}=\sqrt{49}=7. \]
Step 6: Add them.
\[ d_1+d_2=7+7=14. \]
\[ \boxed{14} \]