equation of a chord of an ellipse

Question

There is an ellipse that has equation $x^2+4y^2=36$ and one point inside it $A=(2,1)$. The point is a centre of a chord of an ellipse. I don't know how to find equation of a chord using analytic geometry. Thanks for any ideas how about to start.

Answer 1

I think the trick here is to change coordinates. If you scale one of the axis, two equal length lines will continue equal after that scaling. So, if you take $z=2y$, then the ellipse becomes a circle $x^2 + z^2 = 6^2$, and you point becomes $P=(2,2)$. With that, you can see the chord, in the new coordinates is $z=-x+4$.

So, by returning to the old coordinates it becomes $2y = -x + 4 $.

Answer 2

The equation of the line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$ Substitute for $y$ into the equation of the ellipse and you get a quadratic equation $$x^2(1+4m^2)+x(8m-16m^2)-16m-32=0$$ The roots of the equation are $x_1$ and $x_2$, and the sum of the roots of a quadratic is $-\frac ba$. But we also know that $$\frac {x_1+x_2}{2}=2$$ since $(2,1)$ is the midpoint. Therefore we have $$-\frac{8m-16m^2}{1+4m^2}=4$$ $$\Rightarrow m=-\frac 12$$ Hence you have the equation of the chord.

Answer 3

suppose you parametrize the ellipse by $x = 6\cos t, y = 3 \sin t, 0 \le t \le 2\pi.$ suppose take chord is given by points $t_1, t_2$ and whose center is $A= (2,1).$

then we have $$\begin{align} \cos t_1 + \cos t_2 &= 2/3 \tag 1\\ \sin t_1 + \sin t_2 &= 2/3. \tag 2\end{align}$$

squaring the two equations and subtracting i them gives $$\cos (t_1 + t_2) = 0 \to t_1 + t_2 = \pi/2, 3\pi/2, \ldots$$ subbing $t_2 = \pi/2 - t_1$ in (1) and $(2)$ gives $\sin t_1 + \cos t_1 = 2/3$ but subbing $t_2 = 3\pi/2 - t_1$ in $(1), (2)$ give $\sin t_1 - \cos t_1 = 2/3$ and $\sin t_1 - \cos t_1 = -2/3.$ therefore $$t_1 + t_2=\pi/2. $$

the slope of the chord is $$ \frac{3(\sin t_1 - \sin t_2)}{6(\cos t_1 - \cos t_2)}=-\frac12\frac{\cos (t_1 + t_2)/2}{\sin(t_1+t_2)/2}= -\frac12$$ the equation of the required chord is $$\frac{y-1}{x-2} = -\frac12 \to 2y = -x+4.$$