Analytic Geometry – Problem 2: 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 $(1,\tfrac32)$, and its eccentricity is $\dfrac{\sqrt3}{2}$.

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

(2) Let $A$ be the upper vertex of ellipse $C$. A line $l$ through $T(0,2)$ intersects the ellipse at $M,N$, where $M$ is in the first quadrant. Let the slopes of $AM$, $AN$ be $k_{AM},k_{AN}$, respectively.

(i) Prove that $k_{AM}k_{AN}$ is a constant.

(ii) Through $M$, draw the perpendicular to $AN$, with foot $P$. Let $Q$ be the midpoint of $MP$. Extend $AQ$ to meet $l$ at $G$. Find the range of $\dfrac{TM}{GM}$.

(1) From the eccentricity condition, $\dfrac{b^2}{a^2}=\dfrac14$. Since $(1,\tfrac32)$ lies on the ellipse, $$\frac1{a^2}+\frac{(3/2)^2}{b^2}=1.$$ Solving gives $a^2=4,b^2=1$. Hence $$C:\frac{x^2}{4}+y^2=1.$$

(2) Let $l:y=kx+2\,(k<0)$, and let $M(x_1,y_1),N(x_2,y_2)$. Substitution gives $$(4k^2+1)x^2+16kx+12=0,$$ so $$x_1+x_2=-\frac{16k}{4k^2+1},\qquad x_1x_2=\frac{12}{4k^2+1}.$$

(i) Since $A(0,1)$, $$k_{AM}=\frac{y_1-1}{x_1},\quad k_{AN}=\frac{y_2-1}{x_2}.$$ Substitute $y_i=kx_i+2$ and simplify with Vieta's formulas: $$k_{AM}k_{AN}=\frac1{12}.$$ So it is a constant.

(ii) Following the official derivation, from the equation of $AQ$ and its intersection with $l$, we obtain $$G\left(-\frac3{2k},\frac12\right).$$ Since $y_T=2$, $$\frac{TM}{GM}=\frac{y_T-y_M}{y_M-y_G}=\frac3{2y_M-1}-1.$$ From the intersection constraints, $y_M\in(\tfrac12,1)$, hence $2y_M-1\in(0,1)$, therefore $$\frac{TM}{GM}\in(2,+\infty).$$

(1) The ellipse equation is $$\frac{x^2}{4}+y^2=1.$$ It is determined by combining the eccentricity condition with the fact that $(1,\tfrac32)$ lies on the curve.

(2)(i) The product of slopes is invariant: $$k_{AM}k_{AN}=\frac1{12}.$$ (ii) Using the coordinate relation for $G$ and the admissible interval $y_M\in(\tfrac12,1)$, we get $$\frac{TM}{GM}\in(2,+\infty).$$