Answer
$\frac{x^2}{256} + \frac{y^2}{48}=1$
Work Step by Step
Step 1. Identify the given values: use the diagram given in the Exercise, we can identify a vertex at $(0,16)$ so that $a=16$. We also know that point P$(8,6)$ is on the ellipse.
Step 2. Write a general equation: as the vertex is on the x-axis, we can write an equation as $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ where $a\gt b\gt0$.
Step 3. Find the unknowns: with the above equation and $a=16$, plug-in the coordinate of point P, we have
$\frac{8^2}{16^2} + \frac{6^2}{b^2}=1$ which gives $b^2=48$
Step 4. Conclusion: the equation for the graph can be written as $\frac{x^2}{256} + \frac{y^2}{48}=1$
