4

if the point $A(0,1)$ on the ellipse $\Gamma:$ $\dfrac{x^2}{4}+y^2=1$ and the circle $\tau:$ $(x+1)^2+y^2=r^2(0<r<1)$,if $AB,AC$ tangent the circle $\tau$ ,$B,C\in \Gamma$,show that the line $BC$ always passes through a fixed point

enter image description here

I try Let $AB:y=kx+1$ then $$\begin{cases} \dfrac{x^2}{4}+y^2=1\\ y=kx+1 \end{cases} $$ so we have $$x^2+4(kx+1)^2=4\Longrightarrow (4k^2+1)x^2+8kx=0$$ so we have $$B(-\dfrac{8k}{4k^2+1},\dfrac{1-4k^2}{1+4k^2})$$ But for $C$ it hard to find it.

math110
  • 93,304
  • This problem has a generalization that we can set the ellipse to be any $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the circle with the center $(\pm b,0)$. – m-agag2016 Nov 07 '16 at 11:16
  • @Mingfeng Zhao: You are right, and the center of the circle does not have to exist on the $x$-axis. See the question and the answer here. – mathlove Nov 08 '16 at 08:21

2 Answers2

2

The equation of the line passing through $A(0,1)$ is given by $mx-y+1=0$.

Since we want this line to be tangent to the circle, we have $$r=\frac{|m(-1)-0+1|}{\sqrt{m^2+(-1)^2}},$$ i.e. $$(r^2-1)m^2+2m+r^2-1=0\implies m_1+m_2=\frac{2}{1-r^2},\quad m_1m_2=1\tag1$$

Eliminating $y$ from $mx-y+1=0$ and $x^2/4+y^2=1$ gives $$\frac{x^2}{4}+(mx+1)^2=1\implies x(x+4m^2x+8m)=0\implies x=0,\frac{-8m}{4m^2+1}$$ and $$y=mx+1=m\cdot \frac{-8m}{4m^2+1}+1=\frac{-4m^2+1}{4m^2+1}$$ So we can write $B\left(\frac{-8m_1}{4m_1^2+1},\frac{-4m_1^2+1}{4m_1^2+1}\right),C\left(\frac{-8m_2}{4m_2^2+1},\frac{-4m_2^2+1}{4m_2^2+1}\right)$.

Using $(1)$, we have $$\frac{b_y-c_y}{b_x-c_x}=\frac{\frac{-4m_1^2+1}{4m_1^2+1}-\frac{-4m_2^2+1}{4m_2^2+1}}{\frac{-8m_1}{4m_1^2+1}-\frac{-8m_2}{4m_2^2+1}}=-\frac{m_1+m_2}{3}=\frac{2}{3(r^2-1)}$$and $$\frac{-c_xb_y+b_xc_y}{b_x-c_x}=\frac{-\frac{-8m_2}{4m_2^2+1}\cdot\frac{-4m_1^2+1}{4m_1^2+1}+\frac{-8m_1}{4m_1^2+1}\cdot\frac{-4m_2^2+1}{4m_2^2+1}}{\frac{-8m_1}{4m_1^2+1}-\frac{-8m_2}{4m_2^2+1}}=-\frac{5}{3}$$ So, the equation of the line $BC$ is given by $$y=\frac{b_y-c_y}{b_x-c_x}x+\frac{-c_xb_y+b_xc_y}{b_x-c_x}=\frac{2}{3(r^2-1)}x-\frac 53$$ which passes through $(0,-5/3)$.

mathlove
  • 139,939
  • It's Nice! if we change the point $A(0,1)$ to $A(x_{0},y_{0})$,the $BC$ also passes through a fix point?and How to find it? and I think if we can use geometry methods to explain?Thanks – math110 Nov 04 '16 at 00:15
  • @communnites: I think you can try the method in my answer to solve the version for $A(x_0,y_0)$. Though there might be a geometrical method, I don't know how to solve it geometrically. – mathlove Nov 04 '16 at 06:13
1

When $r\rightarrow 1$, it's easy to see that $y$-axis will be a tangent line, which means that the fixed point should be on the $y$-axis. So later, we need to put $x=0$ in the equation of tangent line to find the fixed $y$-coordinate. Use the tangent conditions, get $\frac{|k-1|}{\sqrt{k^2+1}}=r$ (here, we get $k_1$ and $k_2$), that is, $(1-r^2)k^2-2k+1-r^2=0$, or \begin{eqnarray*} k^2=\frac{r^2+2k-1}{1-r^2}. \end{eqnarray*}

Use Vita's law ($k_1k_2=1$ and $k_1+k_2=\frac{2}{1-r^2}$) to simply the equations of $AB$, you can get \begin{eqnarray*} y=-\frac{2}{3(1-r^2)}(x+\frac{8k_1}{4k_1^2+1})-1+\frac{2}{4k_1^2+1}. \end{eqnarray*}

Set $x=0$, use the first equation to substitute $k^2$ in the above equation and simply, you will get a constant $-\frac{2}{3}-1=-\frac{5}{3}$. So $AB$ must pass through the point $(0,-\frac{5}{3})$.

m-agag2016
  • 934