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An ellipse with the center at $(3,4)$ touches the $x$-axis at $(0,0)$. If the slope of its major axis is $1$, find the eccentricity and equation of the ellipse.

What I try:

Let major axis be $x-y=c$ and minor axis be $x+y=d$. Then $\displaystyle x=\frac{d+c}{2}$ and $\displaystyle y=\frac{d-c}{2}$

So we have $\displaystyle =\frac{d+c}{2}=3$ and $\displaystyle \frac{d-c}{2}=4$

So $\displaystyle d+c=6$ and $d-c=8,$ Solving $d=7,c=-1$

major axis equation is $\displaystyle x-y+1=0$ and minor axis equation is $x+y-7=0$

How do I solve it? Help me, please.

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You have correctly identified the axes. So the equation of the ellipse will be $$\frac{(x+y-7)^2}{a^2}+\frac{(x-y+1)^2}{b^2}=1$$ for some $a,b$. [We have effectively changed coordinates from $x,y$ to $x+y,x-y$.]

We know that it passes through the origin so we need $\frac{49}{a^2}+\frac{7}{b^2}=1$.

Assume $x,y$ are functions of some parameter (eg set $x=a\cos t,y=b\sin t$). Differentiate the equation of the ellipse and set $x=y=\dot{y}=0$ and we get $$\frac{2}{a^2}(-7)(\dot{x})+\frac{2}{b^2}(1)(\dot{x})=0$$ or $a^2=7b^2$. The two equations for $a^2,b^2$ give $a^2=56.b^2=8$.So the equation of the ellipse is:

enter image description here

You can now get the eccentricity from $$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{\frac{6}{7}}$$

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  • $\begingroup$ A very elegant solution :-) $\endgroup$ – Lakshya Sinha Jan 2 at 14:40
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In a modification of the answer by almagest, I explore finding a solution without theuse of differentiation.

My starting point in the first equation derived by almagest:

$\dfrac{(x-y+1)^2}{a^2}+\dfrac{(x+y-7)^2}{b^2}=1$

In this approach put in $y =0$ and derive a quadratic equation for $x$:

$\dfrac{(x+1)^2}{a^2}+\dfrac{(x-7)^2}{b^2}=1$

$(a^2+b^2)x^2-(14a^2-2b^2)x+(49a^2+b^2-a^2b^2)=0$

To get a tangent at $(0,0)$ this equation must have a double root at $x=0$, forcing the linear and constant coefficients both to be zero:

$14a^2-2b^2=0$ Eq. 1

$49a^2+b^2-a^2b^2=0$ Eq. 2

Eq. 1 renders $b^2=7a^2$ and plugging that into Eq. 2 then gives $a^2=8$. Thus

$\dfrac{(x+1)^2}{8}+\dfrac{(x-7)^2}{56}=1$

as found by almagest with his method.

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    $\begingroup$ +1 Yes,you are right. It is probably better to avoid calculus. $\endgroup$ – almagest Jan 2 at 14:50
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    $\begingroup$ +1 indeed, Not trying to be picky but Eq 1. gives $14a^2$ :p you missed out 2 in $2b^2$ ;p $\endgroup$ – Chief VS Jan 2 at 14:54
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    $\begingroup$ Thanks @chief. That is officially my millionth typo on this site. Cheers! :-S $\endgroup$ – Oscar Lanzi Jan 2 at 15:04
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Here’s another way to use the tangency constraint that doesn’t involve calculus. With what you’ve found so far, you know that the equation of the ellipse has the form $${(x+y-7)^2\over a^2}+{(x-y+1)^2\over b^2} = 1.\tag1$$ Note that the equation of the minor axis goes with the major semiaxis length, and vice-versa. The tangent to this ellipse at the point $(x_0,y_0)$ has the equation $${(x+y-7)(x_0+y_0-7)\over a^2}+{(x-y+1)(x_0-y_0+1)\over b^2} = 1.\tag 2$$ Setting $x_0=y_0=0$ in equation (2) and requiring that the coefficient of $x$ vanish (the tangent at the origin has an equation equivalent to $y=0$) produces the equation $${a^2-7b^2\over a^2b^2}=0,\tag3$$ from which $a^2=7b^2$. We can immediately compute the eccentricity: $$\sqrt{1-b^2/a^2}=\sqrt{1-1/7}=\sqrt{6/7}.$$ It’s important to note that this only works because the normals implied by the equations of the principal axes have the same length. This ensures that even though the coefficients $a$ and $b$ in equation (1) aren’t equal to the actual semiaxis lengths of the ellipse, they’re in the same proportion to those lengths.

To find $a^2$ and $b^2$, set $x=y=0$ in equation (1) and use $a^2=7b^2$ to obtain $a^2=56$ and $b^2=8$.

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