Question

A bit out of practice, but how to go about finding the origin and vertix of this equation

9x^(2)+16y^(2)=144

Answer 1

Answer:Explained Below

Step-by-step explanation:

The given equation is similar to an ellipse which is in the form of

$\frac{x^2}{a^2}$+$\frac{y^2}{b^2}$=1

where

2a=length of major axis

2b=length of minor axis

Here after rearranging the given equation we get

$\frac{x^2}{\frac{144}{9}}$+$\frac{y^2}{\frac{144}{16}}$=1

$\frac{x^2}{16}$+$\frac{y^2}{9}$=1

$\frac{x^2}{4^2}$+$\frac{y^2}{3^2}$=1

therefore its origin is (0,0)

and vertices are$\left ( \pm4,0\right )$&$\left ( 0,\pm3\right )$

We can find origin by checking what is with x in the term $\left ( x-something\right )^{2}$

same goes for y

for $\left ( x-2\right )^{2}$ here 2 is the x  coordinate of ellipse

and for vertices Each endpoint of the major axis is vertices and each endpoint of minor axis is co-vertices