Focus of ellipse

Question

If we are given an ellipse with centre (3,4) touches the x axis at (0,0) and if slope of major axis is 1 . Then we have to find the focus of the ellipse .

I tried to rotate the ellipse . But i am not able to proceed .

Can anybody help me in proceeding while rotating the ellipse

Answer 1

How about an answer without rotation?
If you squeeze the whole plane along major axis of the ellipse so that the line x+y=7 is fixed and the image of the ellipse is a circle(blue), then image of the x axis would be a tangent line of that circle, meaning it is orthogonal to the radius.

Image of the origin will stay on the line x=y and on the circle with the diameter AB, A=(3,4) B=(7,0), its equation is $(x-5)^2+(y-2)^2=8$, so its intersection is $$(x-5)^2+(x-2)^2=8$$ No need to find x just to find r of the image of the ellipse, its equation is $$(x-3)^2+(y-4)^2=r^2$$ subtract those equation with y substituted to x you will get $$10x-25+4x-4-6x+9-8x+16=r^2-8$$ having r=2, so the short axis of the ellipse is 2

You can do similar by expanding the plane along short axis so the ellipse go to an circle. you will get system $$(x-1)^2+(-x-2)^2=8$$$$(x-3)^2+(-x-4)^2=R^2$$ So $R^2=28$ which is long axis

Distance of focuses to center is $\sqrt{28-4}=\sqrt{24}$ so $$F=(3±2\sqrt{3},4±2\sqrt{3})$$ Here is the equation of the ellipse $$\frac{(x+y-7)^2}{56}+\frac{(x-y+1)^2}{8}=1$$ I believe you will ask how? well in the center x+y=7 and x-y=-1, and for the most far points on the axises you have $R=\frac{x+y-7}{\sqrt{2}} r=\frac{x-y+1}{\sqrt{2}}$

Answer 2

It is a well known fact that the normal of and ellipse at a point bisects the angle formed by that point and the two foci:

Since the normal at $(0,0)$ to the ellipse is the $y$ axis, we can deduce that the two lines connecting $(0,0)$ and the foci are of the form $y=kx$ and $y=-kx$. We already can deduce that the equation of the major axis is $y=x+1$. Solving for the intersection of $y=kx$ and $y=x+1$, we get the following point: $$\left(\frac{1}{k-1},\frac{k}{k-1}\right)$$ Doing the same for $y=-kx$ and $y=x+1$: $$\left(-\frac{1}{k+1},-\frac{k}{k+1}\right)$$ We know that these points must be equidistant from $(3,4)$, and all three points are collinear, so… $$\frac{1}{k-1}-3=3+\frac{1}{k+1}$$ $$\frac{2}{(k-1)(k+1)}=6$$ $$(k-1)(k+1)=\frac{1}{3}$$ $$k^2-1=\frac{1}{3}$$ $$k^2=\frac{4}{3}$$ $$k=\pm\frac{2\sqrt{3}}{3}$$

So, plugging $k$ back in, we get that the foci are located at $$\left(3+2\sqrt{3},4+2\sqrt{3}\right) \text{ and } \left(3-2\sqrt{3},4-2\sqrt{3}\right)$$

Answer 3

The other answers are good, I just wanted to get the equation in a different way.

Starting with $a x^2 + b x y + c y^2 + d x + e y + f = 0$, right off the bat $f = 0$ since $(0,0)$ is on the curve and $d = 0$ since it's a tangency there and the tangent cone should be $e y$. Now $y = x + 1$ is along the major axis, so we get three more points $(-1,1), (6,8), (7,7) $ on the curve and the linear system:

$a - b + c + e = 0$

$49 a + 49 b +49 c +7e = 0$

$36 a + 48 b + 64 c + 8 e = 0$

We have the freedom to choose one parameter, let's use $e=-7$ to simplify the second equation:

$a - b + c - 7 = 0$

$a + b + c - 1 = 0$

$9 a + 12 b + 16 c - 14 = 0$

Now it's easily got that $a=2, b=-3, c=2$ and we have the equation $2x^2-3x y+2y^2-7y=0$.

Let's make that $4x^2-6x y+4y^2-14y=0$ and apply standard rotation techniques:

$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}4 &-3\\ -3 & 4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}0&-14\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$, and we get eigenvalues 7 and 1 and $P=\frac1{\sqrt{2}}\begin{pmatrix}-1&1\\1&1\end{pmatrix}$, ${\bf x}^tP^t A P{\bf x}+KP{\bf x}=0$. That is $7x'^2+y'^2-7\sqrt{2}x'-7\sqrt{2}y'=0$ and completing squares we get $7(x'-\frac1{\sqrt{2}})^2+(y'-\frac{7}{\sqrt{2}})^2=28$ or $(\frac{x'-\frac1{\sqrt{2}}}{2})^2+(\frac{y'-\frac{7}{\sqrt{2}}}{2\sqrt{7}})^2=1$ Now we have the major axis $2\sqrt{7}$ and the minor axis $2$, so distance from focus to center is $\sqrt{(2\sqrt{7})^2-2^2}=\sqrt{24}$ and the foci are $(3+2\sqrt{3},4+2\sqrt{3})$ and $(3-2\sqrt{3},4-2\sqrt{3})$ from two 45-45-90 triangles with $\sqrt{24}$ as hypotenuse.