Ellipse Problem

[pinkcalculator]

An ellipse is centered at (0,0), passes through (-1,-4) and b is twice as large as a. What is the equation for this ellipse in standard form?

This is as far I've gotten.
b=2a If b>a, the ellipse is vertical

The standard equation is
x^2/a^2 + y^2/b^2 =1

1/a^2 + 16/b^2 =1.

Is this right? Not sure if I've set it up right, or how to finish it.

 

[soroban]

Hello, pinkcalculator!

You have the "twice as large" a bit muddled . . .

An ellipse is centered at (0,0), passes through (-1,-4) and \(\displaystyle b\) is twice as large as \(\displaystyle a.\)
What is the equation for this ellipse in standard form?

\(\displaystyle \text{I will assume that this problem uses the standard equation: }\:\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \;=\;1\)


\(\displaystyle \text{The ellipse is centered at }(0,0)\!:\;h=0,k=0\)

\(\displaystyle \text{We are told that: }\:b \:=\:2a\)
. . . \(\displaystyle \text{The equation becomes: }\:\frac{x^2}{a^2} + \frac{y^2}{(2a)^2} \;=\;1 \quad\Rightarrow\quad \frac{x^2}{a^2} + \frac{y^2}{4a^2} \;=\;1\)

\(\displaystyle \text{The point }(-1,-4)\text{ is on the ellipse: } \;\frac{(-1)^2}{a^2} + \frac{(-4)^2}{4a^2} \;=\;1\)

\(\displaystyle \text{We have: }\;\frac{1}{a^2} + \frac{16}{4a^2} \:=\:1 \quad\Rightarrow\quad \frac{1}{a^2} + \frac{4}{a^2} \:=\:1 \quad\Rightarrow\quad \frac{5}{a^2} \:=\:1 \quad\Rightarrow\quad \boxed{a^2 \:=\:5} \quad\Rightarrow\quad a \:=\:\sqrt{5}\)

. . \(\displaystyle \text{Then: }\:b \:=\:2a \:=\:2\sqrt{5} \quad\Rightarrow\quad\boxed{ b^2 \:=\:20}\)


\(\displaystyle \text{Therefore, the equation of the ellipse is: }\:\frac{x^2}{5} + \frac{y^2}{20} \:=\:1\)

 

[pinkcalculator]

THANK YOU THANK YOU THANK YOU!!
I completely forgot about (h,k) being part of the equation.

I really appreciate your help.