Analytic Geometry – Problem 3: Find the equation of ellipse

Given ellipse $C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$ with eccentricity $\dfrac{\sqrt3}{2}$, and minor-axis length $2$. A line $l$ not passing through the right vertex intersects the ellipse at points $A,B$.

(1) Find the equation of ellipse $C$.

(2) If the circle with diameter $AB$ always passes through the right vertex of the ellipse, determine whether line $l$ passes through a fixed point. If yes, find that fixed point.

(1) Minor-axis length $2\Rightarrow b=1$. Also $e=\dfrac ca=\dfrac{\sqrt3}{2}$. Using $a^2=b^2+c^2$, we obtain $a^2=4$. Therefore $$\frac{x^2}{4}+y^2=1.$$

(2) Let $l:x=ty+m$, and let $A(x_1,y_1),B(x_2,y_2)$. Substituting into the ellipse gives a quadratic in $y$: $$(t^2+4)y^2+2tmy+m^2-4=0,$$ so $$y_1+y_2=-\frac{2tm}{t^2+4},\qquad y_1y_2=\frac{m^2-4}{t^2+4}.$$ Let the right vertex be $M(2,0)$. The condition "$M$ lies on the circle with diameter $AB$" is $$\overrightarrow{MA}\cdot\overrightarrow{MB}=0.$$ Substitute $x_i=ty_i+m$ and the Vieta relations, then simplify to $$5m^2-16m+12=0.$$ Hence $m=\frac65$ or $m=2$. The case $m=2$ means $l$ passes through the right vertex, excluded by the problem condition. Therefore $m=\frac65$, i.e. $$\text{line }l\text{ always passes through the fixed point }\left(\frac65,0\right).$$

(1) The ellipse is $$\frac{x^2}{4}+y^2=1.$$ This follows directly from $b=1$ and $e=\frac{\sqrt3}{2}$.

(2) Yes, $l$ passes through a fixed point. Using the diameter-circle condition $\overrightarrow{MA}\cdot\overrightarrow{MB}=0$, we get the parameter equation $5m^2-16m+12=0$, and after excluding the invalid root, the fixed point is $$\left(\frac65,0\right).$$