2017.3.7 Question 7

as desired, so

lies on the ellipse

.

1. The gradient of $L$ must satisfy that

   $$\begin{array}{llll}\hfill \frac{dy}{dx}& =\frac{dy∕dt}{dx∕dt}& \hfill & \\ \hfill & =\frac{b}{a}\cdot \frac{d\left(2t∕(1+t^2)\right)∕dt}{d\left((1-t^2)∕(1+t^2)\right)∕dt}& \hfill & \\ \hfill & =\frac{b}{a}\cdot \frac{2\cdot (1+t^2)-2t\cdot 2t}{-2t\cdot (1+t^2)-(1-t^2)\cdot 2t}& \hfill & \\ \hfill & =\frac{b}{a}\cdot \frac{2+2t^2-4t^2}{-2t-2t^3-2t+2t^3}& \hfill & \\ \hfill & =\frac{b}{a}\cdot \frac{1-t^2}{-2t}.& \hfill & \end{array}$$

   Therefore, we have a general point $(X,Y)\in L$ satisfy that

   $$\begin{array}{llll}\hfill Y-\frac{2\mathit{bt}}{1+t^2}& =\frac{b}{a}\cdot \frac{1-t^2}{-2t}\cdot \left(X-\frac{a(1-t^2)}{1+t^2}\right)& \hfill & \\ \hfill (1+t^2)Y-2\mathit{bt}& =\frac{b}{a}\cdot \frac{1-t^2}{-2t}\cdot \left((1+t^2)X-a(1-t^2)\right)& \hfill & \\ \hfill (-2\mathit{at})(1+t^2)Y-(-2\mathit{at})(2\mathit{bt})& =b\cdot (1-t^2)\cdot \left((1+t^2)X-a(1-t^2)\right)& \hfill & \\ \hfill (-2\mathit{at})(1+t^2)Y& =b(1-t^2)(1+t^2)X-\mathit{ab}{(1-t^2)}^2-4\mathit{ab}{t}^2& \hfill & \\ \hfill (-2\mathit{at})(1+t^2)Y& =b(1-t^2)(1+t^2)X-\mathit{ab}{(1+t^2)}^2& \hfill & \\ \hfill -2\mathit{atY}& =b(1-t^2)X-\mathit{ab}(1+t^2)& \hfill & \\ \hfill \mathit{ab}(1+t^2)-2\mathit{atY}-b(1-t^2)X& =0& \hfill & \\ \hfill (a+X)b{t}^2-2\mathit{aY}t+b(a-X)& =0& \hfill & \end{array}$$

   as desired.

   Now if we fix $X,Y$ and solve for $t$, there are two solutions to this quadratic equation exactly when

   $$\begin{array}{llll}\hfill {(2\mathit{aY})}^2-4(a+X)b\cdot b(a-X)& >0& \hfill & \\ \hfill {(\mathit{aY})}^2-(a+X)(a-X)b^2& >0& \hfill & \\ \hfill a^2{Y}^2& >(a^2-X^2)b^2,& \hfill & \end{array}$$

   which corresponds to two distinct points on the ellipse.

   Since $a^2{Y}^2>(a^2-X^2)b^2$, we have $\frac{Y^2}{b^2}>1-\frac{X^2}{a^2}$ by dividing through $a^2{b}^2$ on both sides, i.e.

   $$\frac{X^2}{a^2}+\frac{Y^2}{b^2}>1,$$

   which means when the point $(X,Y)$ lies outside the ellipse.

   This also holds when $X^2=a^2$, i.e. when the point $(X,Y)$ lies on the pair of lines $X=\pm A$. Here, the condition is simply $a^2{Y}^2>0$, which gives $Y\ne 0$. One of the tangents will be the vertical line $X=\pm A$ (whichever one the point lies on), and the other one as a non-vertical (as shown when $X=a$, the tangents being $L_1$ and $L_2$).

   

2. By Vieta’s Theorem, we have

   $$\mathit{pq}=\frac{b(a-X)}{b(a+X)}⟹(a+X)\mathit{pq}=a-X,$$

   as desired, and

   $$p+q=-\frac{-2\mathit{aY}}{(a+X)b}=\frac{2\mathit{aY}}{(a+X)b}.$$

   Let $X=0$ for the equation in $L$,

   $$\begin{array}{llll}\hfill \mathit{ab}{t}^2-2\mathit{aY}t+\mathit{ba}& =0& \hfill & \\ \hfill b{t}^2-2Yt+b& =0& \hfill & \\ \hfill Y& =\frac{b(1+t^2)}{2t}.& \hfill & \end{array}$$

   Therefore,

   $$\begin{array}{llll}\hfill y_1+y_2& =\frac{b(1+p^2)}{2p}+\frac{b(1+q^2)}{2q}& \hfill & \\ \hfill & =\frac{b\left[(1+p^2)q+(1+q^2)p\right]}{2\mathit{pq}}& \hfill & \\ \hfill & =2b,& \hfill & \end{array}$$

   therefore we have

   $$4\mathit{pq}=(1+p^2)q+(1+q^2)p=(p+q)(1+\mathit{pq}).$$

   Therefore,

   $$\begin{array}{llll}\hfill 4\cdot \frac{a-X}{a+X}& =\frac{2\mathit{aY}}{(a+X)b}\cdot \frac{2a}{a+X}& \hfill & \\ \hfill a-X& =\frac{a^{2}Y}{b(a+X)}& \hfill & \\ \hfill (a-X)(a+X)b& =a^{2}Y& \hfill & \\ \hfill (a^2-X^2)b& =a^{2}Y& \hfill & \\ \hfill 1-\frac{X^2}{a^2}& =\frac{Y}{b}& \hfill & \\ \hfill \frac{X^2}{a^2}+\frac{Y}{b}& =1,& \hfill & \end{array}$$

   as desired.