Question

Identify the equations of ellipses whose major axis lengths are twice their minor axis lengths.
4x^2+25y^2+32x-250y+589=0

4x^2+y^2+16x+4y+4=0

16x^2+y^2-64x+8y+16=0

2x^2+8y^2-12x+16y-174=0

3x^2+12y^2+18x-24y-69=0

x^2+9y^2-2x+18y-71=0

Answer 1

@welshfella

Answer 2

sorry - ive forgotten this stuff

Answer 3

ok:(

Answer 4

@.Sam.

Answer 5

@perl

Answer 6

you want to find equations that have the two possible forms
$$ \Large{
\frac{(x-h)^2}{(2a)^2} + \frac{(y-k)^2}{a^2}=1

\\~\\ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{(2a)^2}=1
}
$$

Answer 7

how? which is h and which is k?

Answer 8

@perl

Answer 9

h depends on the particular equation, you can complete the square

Answer 10

the long way is to complete the square for each equation

Answer 11

to save time you can plug in x = 0

Answer 12

here is the first graph
https://www.desmos.com/calculator/kedo1nsysm

Answer 13

if you hover your mouse you see that the major axis is length
1 - (-9) = 10
the minor axis is 7 - 3 = 4. and 10 is not twice 4. so that doesnt work

Answer 14

hover your mouse over the vertices

Answer 15

These three functions satisfy youre requirement:
https://www.desmos.com/calculator/yhiapd89td

Answer 16

actually looking at the standard form, we see

Answer 17

$$
\Large{
\frac{(x-h)^2}{(2a)^2} + \frac{(y-k)^2}{a^2}=1
\\~\\\frac{(x-h)^2}{4a^2} + \frac{(y-k)^2}{a^2}=1
\\~\\ 4a^2\left( \frac{(x-h)^2}{4a^2} + \frac{(y-k)^2}{a^2}\right)=4a^2 \cdot 1
\\~\\ 4a^2\cdot \frac{(x-h)^2}{4a^2} + 4a^2 \cdot \frac{(y-k)^2}{a^2})=4a^2 \cdot 1
\\~\\ {(x-h)^2} + 4\cdot(y-k)^2=4a^2
\\~\\ {(x-h)^2} + 4\cdot(y-k)^2 -4a^2=0
}$$

similarly we can show
$$\Large{
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{(2a)^2}=1
\\~\\ \iff \ \\~\\
4(x-h)^2 + (y-k)^2 - 4a^2=0
}
$$