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 $$\displaystyle |a| = \sqrt{5}$$

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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