Find the vertices and foci of the ellipse.

[matt757]

16x²+81y²=81

i already know about the formula i know you have to divide 81 to the left side which will give 16x²/81+y^2/1=1 the part that im confused about is the 16/81 is doesnt divide out like the y part so how do i approach it

 

[pka]

16x²+81y²=81
i already know about the formula i know you have to divide 81 to the left side which will give 16x²/81+y^2/1=1 the part that im confused about is the 16/81 is doesnt divide out like the y part so how do i approach it

Write it as \(\displaystyle \dfrac{x^2}{[(2^{-2})^2(3^2)]^2}+\dfrac{y^2}{1}=1\)

 

[matt757]

why would you write it like that

 

[matt757]

when you use it in the formula c²=a²+b² you would have too for one of them

 

[matt757]

i mean minus on the formula

 

[ksdhart]

You could solve the problem the way pka suggests, but I'll suggest another way which seems far easier to me. Divide both sides of your x-term by 16, like such:

16x² / 81 = x² / (81/16)

Then that leaves you with the standard form you're used to seeing, with a² = 81/16. You should be able to solve the problem from there.