Analytic Geometry – Problem 8: Find the equation of

Given ellipse $C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$, with left and right foci $F_1,F_2$. For any point $P$ on $C$, the perimeter of $\triangle PF_1F_2$ is $6$, and the minimum value of $PF_1$ is $1$.

(1) Find the equation of $C$.

(2) Let $M(4,0)$. A line $l$ through $F_2$ intersects $C$ at $A,B$. Let $k_1,k_2$ be the slopes of $MA$ and $MB$, respectively. Find the range of $k_1k_2$.

(1) For any point $P$ on the ellipse, $$PF_1+PF_2=2a.$$ The perimeter condition gives $2a+2c=6$, so $$a+c=3.$$ Given that the minimum of $PF_1$ is 1 and $$PF_1^{\min}=a-c=1,$$ we have $$a+c=3,\quad a-c=1.$$ Solving gives $a=2,c=1$, therefore $$b^2=a^2-c^2=3.$$ So $$\frac{x^2}{4}+\frac{y^2}{3}=1.$$

(2) Split into two cases:

- If $l$ coincides with the $x$-axis, then $A,B$ lie on the $x$-axis, so $k_1=k_2=0$, hence $k_1k_2=0$.

- If $l$ is not the $x$-axis, set $$l:x=my+1.$$ Let intersections be $A(x_1,y_1),B(x_2,y_2)$. Solving $$\begin{cases} x=my+1,\\ 3x^2+4y^2=12 \end{cases}$$ gives $$(3m^2+4)y^2+6my-9=0,$$ so $$y_1+y_2=-\frac{6m}{3m^2+4},\qquad y_1y_2=-\frac{9}{3m^2+4}.$$ Also, $$k_1=\frac{y_1}{x_1-4}=\frac{y_1}{my_1-3}, \quad k_2=\frac{y_2}{x_2-4}=\frac{y_2}{my_2-3}.$$ Substituting and simplifying gives $$k_1k_2=-\frac{1}{4(m^2+1)}.$$ Since $m\in\mathbb R$, $$-\frac14\le k_1k_2<0.$$

Combining both cases, $$k_1k_2\in\left[-\frac14,0\right].$$

(1) The ellipse is $$\frac{x^2}{4}+\frac{y^2}{3}=1.$$ This is obtained from $a+c=3$ and $a-c=1$, which come from the perimeter and minimum-distance conditions.

(2) For non-horizontal lines through $F_2$, we derive $k_1k_2=-\dfrac{1}{4(m^2+1)}\in\left[-\dfrac14,0\right)$. Including the special horizontal case gives $k_1k_2=0$, so finally $$k_1k_2\in\left[-\frac14,0\right].$$