Use the discriminant to classify the graph; then use the quadratic formula to solve for $y$.
$$3 x ^ { 2 } - 2 \sqrt { 3 } x y + y ^ { 2 } - 18 \sqrt { 3 } x + 30 y + 36 = 0$$
A) ellipse; $$y = - 15 + \sqrt { 3 } x \pm \sqrt { 12 \sqrt { 3 } x + 189 }$$
B) ellipse; $y = - 15 + \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$
C) parabola; $y = - 15 + \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$

D) parabola; $y = \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$

E) hyperbola;$y = \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$

Question

Use the discriminant to classify the graph; then use the quadratic formula to solve for $y$.
$$3 x ^ { 2 } - 2 \sqrt { 3 } x y + y ^ { 2 } - 18 \sqrt { 3 } x + 30 y + 36 = 0$$
A) ellipse; $$y = - 15 + \sqrt { 3 } x \pm \sqrt { 12 \sqrt { 3 } x + 189 }$$
B) ellipse; $y = - 15 + \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$
C) parabola; $y = - 15 + \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$

D) parabola; $y = \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$

E) hyperbola;$y = \sqrt { 3 } x \pm \sqrt { - 12 \sqrt { 3 } x + 189 }$

Quizplus · Accepted Answer

The discriminant of this quadratic is $\Delta = b^2 - 4ac = (-2\sqrt{3})^2 - 4(3)(1) = -24 < 0$. This tells us that the graph is a parabola. Using the quadratic formula, we can solve for $y$:
\begin{align*}
3x^2 - 2\sqrt{3}xy + y^2 -18\sqrt{3}x +30y + 36 &= 0 \
y^2 - 2\sqrt{3}xy + 3x^2 + 30y - 18\sqrt{3}x + 36 &= 0 \
y &= \frac{ \pm \sqrt{(-2\sqrt{3}x)^2 - 4(3)(3x^2 + 30y - 18\sqrt{3}x + 36) }}{2} \
&= -15 + \sqrt{3}x \pm \sqrt{-12\sqrt{3}x+189}
\end{align*} Therefore, the answer is choice C.

Use the Discriminant to Classify the Graph; Then Use the Quadratic