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

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.

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

• A very elegant solution :-) – Lakshya Sinha Jan 2 at 14:40

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.

• +1 Yes,you are right. It is probably better to avoid calculus. – almagest Jan 2 at 14:50
• +1 indeed, Not trying to be picky but Eq 1. gives $14a^2$ :p you missed out 2 in $2b^2$ ;p – Chief VS Jan 2 at 14:54
• Thanks @chief. That is officially my millionth typo on this site. Cheers! :-S – Oscar Lanzi Jan 2 at 15:04

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 $$b^2=8$$.