Q.The ellipse $E_1: \dfrac{x^2}9+\dfrac{y^2}4=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse is

Setting the dot product of $\vec{BF}$ and $\vec{BF'}$ to zero (right angle at $B$) and combining with the ellipse identity $b^2=a^2(1-e^2)$ gives $e=1/\sqrt2$.

1. Let the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$). $B=(0,b)$ is the end of the semi-minor axis; the foci are $F=(ae,0)$, $F'=(-ae,0)$.
2. $\vec{BF}=(ae-0,\,0-b)=(ae,-b)$; $\vec{BF'}=(-ae-0,\,0-b)=(-ae,-b)$.
3. Angle $FBF'=90^\circ$ means $\vec{BF}\cdot\vec{BF'}=0$: $(ae)(-ae)+(-b)(-b)=0\Rightarrow-a^2e^2+b^2=0\Rightarrow b^2=a^2e^2$. …

Converts to standard ellipse form to read off $a$, then applies the defining focal-distance property $PF_1+PF_2=2a$.

1. Divide $16x^2+25y^2=400$ by $400$: $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$.
2. So $a^2=25\Rightarrow a=5$ (since $25>16$, the major axis is along $x$, matching foci on the $x$-axis at $(\pm3,0)$). …

Since $(\pm\sqrt5,0)$ are the foci of the ellipse $4x^2+9y^2=36$, the constant sum of focal distances from any point on the ellipse is $2a=6$.

1. Rewrite $4x^2+9y^2=36$ in standard form by dividing by $36$: $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$.
2. Here $a^2=9$ (larger denominator, under $x^2$) and $b^2=4$, so $a=3,\ b=2$.
3. The distance of each focus from the center is $c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt5$. …

Complete the square in $x$ and $y$ to bring the conic to standard ellipse form, identify the larger denominator as $a^2$, and apply $e=\sqrt{1-b^2/a^2}$.

1. Given: $9x^2+5y^2-54x-40y+116=0$.
2. Group and complete the square: $9(x^2-6x) + 5(y^2-8y) + 116 = 0$.
3. $9(x^2-6x+9-9) + 5(y^2-8y+16-16) + 116=0 \Rightarrow 9(x-3)^2-81 + 5(y-4)^2-80+116=0$.
4. Simplify constants: $-81-80+116=-45$, so $9(x-3)^2+5(y-4)^2-45=0 \Rightarrow 9(x-3)^2+5(y-4)^2=45$.
5. Divide by 45: $\dfrac{(x-3)^2}{5}+\dfrac{(y-4)^2}{9}=1$. …