# Two points on an elipse; trying to find foci and vertex/minor axis

#### smash

Hi there new here. This seems like a really easy question but can't get it:

basically standard form of ellipse:

Code:
x2     y2
--  +  --  = 1
a2     b2
Here's the question: ''pass through (2,2) and (1,4)''

plz solve, thanks!

(what I did: filled in two of these equations with this info on top and two sets of unknown a's and b's, but have no idea where to take it after that.)

#### bigwave

maybe I am misunderstanding your question but any number of elipses could pass through 2 given points we would need more info

besides this. welcome to the forum, it is a great place to be.

Last edited:

#### earboth

MHF Hall of Honor
Hi there new here. This seems like a really easy question but can't get it:

basically standard form of ellipse:

Code:
x2     y2
--  +  --  = 1
a2     b2
Here's the question: ''pass through (2,2) and (1,4)''

plz solve, thanks!

(what I did: filled in two of these equations with this info on top and two sets of unknown a's and b's, but have no idea where to take it after that.)
1. I assume that you already have:

$$\displaystyle \left|\begin{array}{rcl}\frac4{a^2}+\frac4{b^2} &=&1 \\ \frac1{a^2}+\frac{16}{b^2} &=&1\end{array}\right.$$ which yields: $$\displaystyle \left|\begin{array}{rcl}4b^2+4a^2 &=&a^2 b^2 \\ b^2+16a^2&=& a^2 b^2\end{array}\right.$$

2. Subtract the 2nd equation from the 1st one and you'll get:

$$\displaystyle b^2 = 4a^2~\implies~|b| = 2\cdot |a|$$

3. Plug in this value into the 1st (or 2nd) equation. You'll get $$\displaystyle |a| = \sqrt{5}$$

#### Attachments

• 16.4 KB Views: 18

#### HallsofIvy

MHF Helper
maybe I am misunderstanding your question but any number of elipses could pass through 2 given points we would need more info
Yes, but the given form, $$\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}$$ implies that the axes of symmetry are on the x and y axes and the center is at (0, 0) which reduces it to only one ellipse.

besides this. welcome to the forum, it is a great place to be.

• bigwave