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Context:

  1. Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(1,\sqrt{6})$ and $(3,0)$.
  2. Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(2,4)$ and $(5,\sqrt{2})$.

In problems like these, I frequently find myself solving simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$. For example, when doing math 2, I had to solve the following system of equations (I assumed $p=a^2$, $q=b^2$).

$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$

$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$

I'll now show how I solved them.

$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$

$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$

Now in $(1)$,

$$\frac{4q+16p}{pq}=1$$

$$4q+16p=pq$$

Again in $(2)$,

$$\frac{25}{p}+\frac{2}{q}$$

$$\frac{25q+2p}{pq}=1$$

$$25q+2p=pq$$

Now,

$$4q+16p=25q+2p$$

$$21q-14p=0$$

$$21q=14p$$

$$q=\frac{14p}{21}$$

Now,

$$\frac{4}{p}+\frac{16}{q}=1$$

$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$

$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$

$$\frac{4}{p}+\frac{336}{14p}=1$$

$$\frac{56+336}{14p}=1$$

$$14p=392$$

$$p=28$$

Now,

$$\frac{4}{28}+\frac{16}{q}=1$$

$$\frac{1}{7}+\frac{16}{q}=1$$

$$\frac{16}{q}=\frac{6}{7}$$

$$6q=112$$

$$q=\frac{56}{3}$$

So, the equation of the ellipse is $$\frac{x^2}{28}+\frac{y^2}{\frac{56}{3}}=1\ \text{(Ans.)}$$

As you can see, the process is pretty long and time-taking. Is there any way to find the values of $p,q$ in 2-3 lines?

tryingtobeastoic
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    Repeat the same procedure you know how to do for an arbitrary a and b. Then you will have a formula you can just plugin a and b and get the answer :) – Nurator Feb 22 '22 at 06:40
  • @Nurator Yeah, that's a good point. I'll still have to remember the formula though, which is a disadvantage. Is there another way? – tryingtobeastoic Feb 22 '22 at 06:43
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    In your example, you just have two linear equations in variables $p^{-1}$ and $q^{-1}$. Solve for $p^{-1}$ and $q^{-1}$ using elimination or substitution, and deduce the values of $p$ and $q$. – Theo Bendit Feb 22 '22 at 06:44
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    Regarding what @Theo Bendit said, see this answer. – Dave L. Renfro Feb 22 '22 at 08:45
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    @tryingtobeastoic To make your life easier, let $\alpha=\dfrac{1}{a^2},\beta=\dfrac{1}{b^2}$ and now just need to solve for simultaneous linear equations only. – Ng Chung Tak Feb 22 '22 at 22:45

2 Answers2

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I've found a quicker way:

$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$

$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$

Now, $(2)\times8-(1)$:

$$\frac{200}{p}+\frac{16}{q}-\frac{4}{p}-\frac{16}{q}=8-1$$

$$\frac{196}{p}=7$$

$$p=\frac{196}{7}$$

$$p=28$$

Now, inputting $p=28$ in $(1)$,

$$\frac{4}{28}+\frac{16}{q}=1$$

$$\frac{16}{q}=\frac{6}{7}$$

$$q=7\cdot\frac{16}{6}$$

$$q=\frac{56}{3}$$

$$(p,q)=(28,\frac{56}{3})$$

tryingtobeastoic
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Use the substitution $x = \dfrac{1}{p} , y = \dfrac{1}{q} $, then your equations become:

$ 4 x + 16 y = 1 $

$ 25 x + 2 y = 1 $

And these can solved very easily, using for example, elimination

$(1) - 8 \times (2) $:

$- 196 x = - 7 \Rightarrow x = \dfrac{7}{196} \Rightarrow p = \dfrac{196}{7} = 28 $

And then, from equation $(2)$,

$y = \dfrac{1}{2} ( 1 - 25 x ) = \dfrac{3}{56} $

Therefore, $q = \dfrac{56}{3}$

Hosam Hajeer
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