Curve Bounding: x^2/16 + y^2/9 = 1

[skatru]

I need help finding the function y = f(x) that gives the curve bounding the top of the ellipse.

The equation of the ellipse is x^2/16 + y^2/9 = 1

I'm just not sure even where to start.

Thanks for any help.

 

[skeeter]

Re: Curve Bounding

solve for y ... the (+) square root of y as a function of x is the top half of the ellipse.

 

[skatru]

the square root of -9(x^2/19 - 1)

Is that right?

 

[galactus]

No, not quite.

\(\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)

The LCM of 9 and 16 is 144, so multiply through by 144 and get:

\(\displaystyle 9x^{2}+16y^{2}=144\)

\(\displaystyle 16y^{2}=144-9x^{2}\)

\(\displaystyle y^{2}=\frac{144-9x^{2}}{16}\)

\(\displaystyle y=\sqrt{\frac{144-9x^{2}}{16}}\)

Factor out a 9 in the radical:

\(\displaystyle y=\sqrt{\frac{9(16-x^{2})}{16}}\)

Bring out the \(\displaystyle \sqrt{\frac{9}{16}}=\frac{3}{4}\)

\(\displaystyle y=\pm\frac{3}{4}\sqrt{16-x^{2}}, \;\ and \;\ 16-x^{2}\geq{0}\)

See?.