Analytic Geometry – Problem 18: Find the fixed point on the -axis through which line always passes

Given ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$ with left and right foci $F_1,F_2$, the focus of parabola $y^2=4x$ coincides with $F_2$. Let $P$ be the intersection point of the ellipse and parabola in the first quadrant, and $PF_1=\dfrac73$. Through $F_2$, draw two nonzero-slope perpendicular lines: one intersects the ellipse at $A,B$, and the other intersects the ellipse at $C,D$. Let $M$ be the midpoint of chord $AB$, and $N$ be the midpoint of chord $CD$. Find the fixed point on the $x$-axis through which line $MN$ always passes.

The parabola is $y^2=4x$, so its focus is $F_2(1,0)$. Hence the ellipse right focus is also $F_2$, giving $$c=1.$$ Let $P(x_P,y_P)$ be the common point in the first quadrant. From $PF_1=\dfrac73$ and $F_1(-1,0)$: $$(x_P+1)^2+y_P^2=\frac{49}{9}.$$ Using $y_P^2=4x_P$, $$x_P^2+6x_P+1=\frac{49}{9}\Rightarrow 9x_P^2+54x_P-40=0,$$ so $x_P=\dfrac23$ (reject negative root), and $y_P^2=\dfrac83$.

Now $$PF_2=\sqrt{(x_P-1)^2+y_P^2}=\sqrt{\frac19+\frac83}=\frac53.$$ For ellipse, $PF_1+PF_2=2a$, so $$2a=\frac73+\frac53=4\Rightarrow a=2.$$ Then $$b^2=a^2-c^2=4-1=3,$$ hence the ellipse is $$\frac{x^2}{4}+\frac{y^2}{3}=1.$$

Take a line through $F_2(1,0)$: $y=m(x-1)$, meeting ellipse at $A,B$. From Vieta on $$\frac{x^2}{4}+\frac{m^2(x-1)^2}{3}=1,$$ the midpoint $M$ of $AB$ is $$x_M=\frac{4m^2}{3+4m^2},\qquad y_M=-\frac{3m}{3+4m^2}.$$ For the perpendicular line $y=-\dfrac1m(x-1)$, with intersections $C,D$, midpoint $N$ is $$x_N=\frac{4}{3m^2+4},\qquad y_N=\frac{3m}{3m^2+4}.$$ Write the two-point equation of line $MN$; setting $y=0$ gives the $x$-intercept $$x=\frac47,$$ independent of $m$. Therefore $MN$ always passes through $$\left(\frac47,0\right).$$

$$MN\text{ always passes through }\left(\frac47,0\right).$$