1. Identify the conic

The given equation is $9x^2+16y^2=144.$ Divide by $144$: $\frac{x^2}{16}+\frac{y^2}{9}=1.$

This is an ellipse in standard form: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a^2=16,\quad b^2=9.$ So, $a=4,\quad b=3.$

2. Find the foci of the ellipse

For an ellipse, $c^2=a^2-b^2=16-9=7.$ Hence, $c=\sqrt{7}.$

Therefore the foci are $(\pm \sqrt{7},0).$

Given points are: $A=(\sqrt7,0),\qquad B=(-\sqrt7,0).$ So $A$ and $B$ are exactly the two foci of the ellipse.

3. Use the property of ellipse

For any point $P$ on an ellipse, the sum of distances from the two foci is constant and equal to $2a$. Thus, $PA+PB=2a=2\times 4=8.$

4. Check options

• A: $8$ ✅
• B: $9$
• C: $16$
• D: $6$

So the correct option is A.