AP DSC School Assistant Math Model Test 2

Concept:An ellipse is represented by a second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ when the discriminant $B^2 - 4AC < 0$ and $A$ and $C$ have the same sign. Additionally, the equation must be transformable to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with positive denominators.Explanation:Given equation: $\frac{x^{2}}{2-r} + \frac{y^{2}}{r-6} + 1 = 0$.First, rewrite it as $\frac{x^{2}}{2-r} + \frac{y^{2}}{r-6} = -1$. Multiply both sides by $-1$ to get $\frac{x^{2}}{r-2} + \frac{y^{2}}{6-r} = 1$. For this to represent an ellipse, both denominators must be positive:- $r-2 > 0 \implies r > 2$,- $6-r > 0 \implies r < 6$.Thus $2 < r < 6$.Alternatively, using the discriminant condition: Compare with $Ax^2 + Bxy + Cy^2 + \dots = 0$. Here $A = \frac{1}{2-r}$, $B = 0$, $C = \frac{1}{r-6}$. The condition $B^2 - 4AC < 0$ gives:$$0^2 - 4\left(\frac{1}{2-r}\right)\left(\frac{1}{r-6}\right) < 0 \implies -4\frac{1}{(2-r)(r-6)} < 0 \implies \frac{1}{(2-r)(r-6)} > 0.$$Since numerator $1>0$, denominator must be positive: $(2-r)(r-6) > 0$. Multiply by $-1$: $(r-2)(r-6) < 0$, which implies $r$ lies between $2$ and $6$, i.e., $2 < r < 6$.Additionally, $A \neq C$ is automatically satisfied for $r$ in this interval.Answer:Option C: $2 < r < 6$