Analytic Geometry – Problem 1: Find the standard equation of ellipse

Given the ellipse $C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$, which passes through $P(\tfrac32,3)$ and $Q(0,2)$. A line $l$ through $(0,1)$ intersects the ellipse $C$ at points $A,B$. Let $O$ be the origin.

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

(2) If the area of $\triangle QAB$ is $\tfrac{3\sqrt3}{2}$, find the equation of line $l$.

(3) Point $M$ lies on ray $OB$. If there exists a point $N$ on the ellipse such that quadrilateral $OANM$ is a parallelogram, find the range of $\dfrac{OM}{OB}$.

(1) Substitute $P,Q$ into the ellipse equation: $$\frac{(3/2)^2}{a^2}+\frac{3^2}{b^2}=1,\qquad \frac{2^2}{b^2}=1.$$ Hence $b^2=4$, and then $a^2=9$. Therefore $$C:\frac{x^2}{9}+\frac{y^2}{4}=1.$$

(2) Let $l:y=kx+1$, and let $A(x_1,y_1),B(x_2,y_2)$. Solve $$\begin{cases} y=kx+1,\\ 4x^2+9y^2=36. \end{cases}$$ Eliminating $y$ gives $$(4+9k^2)x^2+18kx-27=0.$$ By Vieta's formulas, we obtain $x_1+x_2$ and $x_1x_2$. Using $$S_{\triangle QAB}=\frac12\cdot1\cdot|x_1-x_2|=\frac{3\sqrt3}{2}.$$ we simplify to $k=0$. Thus $$l:y=1.$$

(3) Let $\overrightarrow{OM}=m\overrightarrow{OB}$. Then $$M=(mx_2,my_2),\quad N=(x_1+mx_2,\,y_1+my_2).$$ Since $A,B,N$ are on the ellipse, simplification gives $$4x_1x_2+9y_1y_2=-18m.$$ Combining this with the Vieta relations from part (2), we get $$m=2-\frac4{4+9k^2}.$$ Because $k\in\mathbb R$, $$m\in[1,2).$$ So $$\frac{OM}{OB}\in[1,2).$$

(1) The ellipse is $$\frac{x^2}{9}+\frac{y^2}{4}=1.$$ This comes from solving $a^2,b^2$ using the two given points on the curve.

(2) Using the area condition $S_{\triangle QAB}=\tfrac{3\sqrt3}{2}$, the slope parameter is $k=0$, so the line is $l:y=1$.

(3) Writing $\overrightarrow{OM}=m\overrightarrow{OB}$ and imposing the parallelogram condition with a point $N$ on the ellipse yields $m\in[1,2)$. Therefore $$\frac{OM}{OB}\in[1,2).$$