Graphical representation of an equation

Question

How to plot a graph of the equation $12x^2-36x +15+ 16y^2 = 0$ I don't know how to plot that. Can you please give me the graph of this equation?

Answer 1

You have an elipse generally of the form $$\frac{(X-X_0)^2}{a^2}+\frac{(Y-Y_0)^2}{b^2}=1$$ where $(X_0,Y_0)$ is the center of the ellipse, and a and b are the semi-axes. Somehow you have to get your problem into this form. But immediately you see that there is no Y term, so you know that $Y_0 = 0$.

Knowing that the widest diameter of the ellipse is on the X axis, you can solve as Leibovici above did.

To find the left and right boundaries of the ellipse, set Y=0 and factor $12X^2-36X+15$ to get $3(2x-5)(2x-1)$. The center of the ellipse is half way between these two roots of $\frac{5}{2}$ and $\frac{1}{2}$ at $X_0 = \frac{3}{2}$.

To find the top and bottom boundaries of the ellipse, set $X = \frac{3}{2}$ and solve for Y.

The bottleneck in the problem was the factoring job. Once that was done, the rest was easy!

Answer 2

Observe we have \begin{align} 12x^2-36x+15+16y^2=&\ 12(x^2-3x+\frac{9}{4})-27+15+16y^2\\ =&\ 12\left(x-\frac{3}{2}\right)^2+16y^2-14=0 \end{align} which means \begin{align} \frac{\left(x-\frac{3}{2}\right)^2}{14/12}+\frac{y^2}{14/16} = 1 \end{align} i.e. you have an ellipse centered at $(3/2, 0)$.

Answer 3

You can make the problem explicit solving for $y$ equation $$12x^2-36x +15+ 16y^2 = 0$$ This will give $$y=\pm \frac{\sqrt{3}}{4} \sqrt{-4 x^2+12 x-5}$$ Since $-4 x^2+12 x-5 \geq 0\implies \frac 12 \leq x \leq \frac 52$.

Plot the two branches and get the ellipse identified by Jacky Chong.