# How do you identity if the equation 25y^2+9x^2-50y-54x=119 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

May 31, 2018

The conic is an ellipse

#### Explanation:

We have:

$25 {y}^{2} + 9 {x}^{2} - 50 y - 54 x = 119$

First we can collect "like" terms:

$\left\{9 {x}^{2} - 54 x\right\} + \left\{25 {y}^{2} - 50 y\right\} = 119$

Next we complete the square independently on the $x$ and $y$ terms

$\therefore \left\{9 \left({x}^{2} - \frac{54}{9} x\right)\right\} + \left\{25 \left({y}^{2} - \frac{50}{25} y\right)\right\} = 119$

$\therefore 9 \left\{{x}^{2} - 6 x\right\} + 25 \left\{{y}^{2} - 2 y\right\} = 119$

$\therefore 9 \left\{{\left(x - 3\right)}^{2} - {3}^{2}\right\} + 25 \left\{{\left(y - 1\right)}^{2} - {1}^{2}\right\} = 119$

$\therefore 9 \left\{{\left(x - 3\right)}^{2} - 9\right\} + 25 \left\{{\left(y - 1\right)}^{2} - 1\right\} = 119$

We have now factorised the "quadratic" terms so we now collect constant terms:

$\therefore 9 {\left(x - 3\right)}^{2} - 9 \cdot 9 + 25 {\left(y - 1\right)}^{2} - 25 \cdot 1 = 119$

$\therefore 9 {\left(x - 3\right)}^{2} + 25 {\left(y - 1\right)}^{2} = 119 + 25 + 81$

$\therefore 9 {\left(x - 3\right)}^{2} + 25 {\left(y - 1\right)}^{2} = 225$

Finally, we put the equation into standard form:

$\therefore \frac{9}{225} {\left(x - 3\right)}^{2} + \frac{25}{225} {\left(y - 1\right)}^{2} = 1$

$\therefore \frac{1}{25} {\left(x - 3\right)}^{2} + \frac{1}{9} {\left(y - 1\right)}^{2} = 1$

$\therefore {\left(x - 3\right)}^{2} / {5}^{2} + {\left(y - 1\right)}^{2} / {3}^{2} = 1$

And as such, we can identify the conic as an ellipse

graph{25y^2 + 9x^2 - 50y - 54x = 119 [-10, 10, -5, 5]}