Step 1: Identify what kind of curve we are looking for.
The graph opens sideways (left and right) rather than up and down, so we need an equation whose standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$: a hyperbola opening along the x-axis, with vertices at $(\pm a, 0)$.
Step 2: Convert each quadratic option into this standard form.
Option 1: $8x^2-15y^2=169 \Rightarrow \frac{x^2}{169/8}-\frac{y^2}{169/15}=1$, so $a=\sqrt{169/8}\approx4.6$.
Option 2: $9x^2-16y^2=144 \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$, so $a=\sqrt{16}=4$ exactly.
Options 3 and 4 are not conics of this type at all; they involve $|(x-h)(y-k)|=c$, a curve centred away from the origin, not symmetric about both axes the way the picture is.
Step 3: Match against the picture.
The graph is symmetric about both the x-axis and the y-axis, and its vertices sit cleanly between the gridlines marked 2 and 6, consistent with $a=4$ exactly. Option 1's vertex near 4.6 does not line up as cleanly, and options 3 and 4 are not even centred at the origin.
Step 4: Final Answer.
The equation that fits is $9x^2-16y^2=144$.
\[ \boxed{9x^2-16y^2=144} \]
