What are the conic sections of the following equations 16x^2 + 25y^2- 18x - 20y + 8=0?

May 16, 2018

It is an ellipse.

Explanation:

The above equation can be easily converted into the ellipse form ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ as coefficients of ${x}^{2}$ and${y}^{2}$ both are positive), where $\left(h , k\right)$ is the center of ellipse and axis are $2 a$ and $2 b$, with larger one as major axis an other minor axis. We can also find vertices by adding $\pm a$ to $h$ (keeping ordinate same) and $\pm b$ to $k$ (keeping abscissa same).

We can write the equation $16 {x}^{2} + 25 {y}^{2} - 18 x - 20 y + 8 = 0$ as

$16 \left({x}^{2} - \frac{18}{16} x\right) + 25 \left({y}^{2} - \frac{20}{25} y\right) = - 8$

or $16 \left({x}^{2} - 2 \cdot \frac{9}{16} x + {\left(\frac{9}{16}\right)}^{2}\right) + 25 \left({y}^{2} - 2 \cdot \frac{2}{5} y + {\left(\frac{2}{5}\right)}^{2}\right) = - 8 + 16 {\left(\frac{9}{16}\right)}^{2} + 25 {\left(\frac{2}{5}\right)}^{2}$

or $16 {\left(x - \frac{9}{16}\right)}^{2} + 25 {\left(y - \frac{2}{5}\right)}^{2} = - 8 + \frac{81}{16} + 4$

or $16 {\left(x - \frac{9}{16}\right)}^{2} + 25 {\left(y - \frac{2}{5}\right)}^{2} = \frac{17}{16}$

or ${\left(x - \frac{9}{16}\right)}^{2} / {\left(\frac{\sqrt{17}}{16}\right)}^{2} + {\left(y - \frac{2}{5}\right)}^{2} / {\left(\frac{\sqrt{17}}{20}\right)}^{2} = 1$

Hence center of ellipse is $\left(\frac{9}{16} , \frac{2}{5}\right)$, while major axis parallel to $x$-axis is $\frac{\sqrt{17}}{8}$ and minor axis parallel to $y$-axis is $\frac{\sqrt{17}}{10}$.

graph{(16x^2+25y^2-18x-20y+8)((x-9/16)^2+(y-2/5)^2-0.0001)(x-9/16)(y-2/5)=0 [-0.0684, 1.1816, 0.085, 0.71]}