Analytic Geometry – Problem 4: Find the standard equation of ellipse

As shown in the figure, ellipse $C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$ has eccentricity $\dfrac{\sqrt2}{2}$. Line $l_1$ passes through $A(-6,0)$ and the upper vertex of the ellipse, and its slope is $\dfrac{\sqrt3}{3}$.

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

(2) Let line $l_2:y=x+t$ intersect the ellipse at $P,Q$ (with $Q$ below $P$). The second intersection of line $AP$ with the ellipse is $M$, and the second intersection of line $AQ$ with the ellipse is $N$.

(i) When $N$ coincides with $Q$, find $t$.

(ii) Prove that as $t$ varies, line $MN$ passes through a fixed point.

(1) From $e=\dfrac ca=\dfrac{\sqrt2}{2}$, we get $b=c$. Since $l_1$ passes through $(-6,0)$ and the upper vertex $(0,b)$, $$\frac{b-0}{0-(-6)}=\frac{\sqrt3}{3}\Rightarrow b=2\sqrt3.$$ Therefore $c=2\sqrt3$, and $$a^2=b^2+c^2=24,\quad b^2=12.$$ Thus $$\frac{x^2}{24}+\frac{y^2}{12}=1.$$

(2)(i) When $N\equiv Q$, line $AQ$ is tangent to the ellipse. Let $AQ:x=ny-6\,(n<0)$. Intersecting with the ellipse and setting the discriminant to zero gives $n=-1$, so the tangency point is $$Q(-4,-2).$$ Substitute into $l_2:y=x+t$: $-2=-4+t\Rightarrow t=2$.

(2)(ii) Let $M(x_1,y_1),N(x_2,y_2)$. Following the official Vieta-based derivation for the second intersections on $AM$, $AN$, $$P\left(\frac1{x_1+5}-5,\frac{y_1}{x_1+5}\right), \quad Q\left(\frac1{x_2+5}-5,\frac{y_2}{x_2+5}\right).$$ Using $k_{PQ}=1$, we simplify to $$m-5k=1$$ for the line $MN:y=kx+m$. Hence $$y=kx+m=k(x+5)+1,$$ so $MN$ always passes through the fixed point $$(-5,1).$$

(1) The standard equation is $$\frac{x^2}{24}+\frac{y^2}{12}=1.$$ It is obtained from $e=\frac{\sqrt2}{2}$ and the slope condition of the line through $(-6,0)$ and the top vertex.

(2)(i) In the tangent case $N\equiv Q$, solving the tangency condition gives $Q(-4,-2)$, so $t=2$.

(2)(ii) The relation $m-5k=1$ for $MN:y=kx+m$ implies $y=k(x+5)+1$, hence every such line $MN$ passes through the fixed point $(-5,1)$.