[Math] 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.

analytic geometryconic sections

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.

Best Answer

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:

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

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