Q.Find the equation of the ellipse in each of the cases given below:

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 $\vec{BF}\cdot\vec{BF'}=0$: $(ae)(-ae)+(-b)(-b)=0\Rightarrow-a^2e^2+b^2=0\Rightarrow b^2=a^2e^2$.
4. Also, for any ellipse, $b^2=a^2(1-e^2)$ (the standard relation between semi-axes and eccentricity).
5. Equate the two expressions for $b^2$: $a^2(1-e^2)=a^2e^2\Rightarrow1-e^2=e^2\Rightarrow1=2e^2\Rightarrow e^2=\dfrac12$.
6. So $e=\dfrac{1}{\sqrt2}$ (taking the positive root, since eccentricity is positive).

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 $(\pm3,0)$).
3. By the defining property of an ellipse, for any point $P$ on it, the sum of distances to the two foci is constant and equal to $2a$: $PF_1+PF_2=2(5)=10$.

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 $c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt5$.
4. So the foci are at $(\sqrt5,0)$ and $(-\sqrt5,0)$ — exactly the two points given in the question.
5. By the defining property of an ellipse, the sum of the distances from any point on the ellipse to the two foci is the constant $2a$.
6. Here $2a=2\times3=6$.

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: $\dfrac{(x-3)^2}{5}+\dfrac{(y-4)^2}{9}=1$.
6. Since $9>5$, the major axis is along the $y$-direction, so $a^2=9$ (under the $y$ term) and $b^2=5$ (under the $x$ term).
7. Eccentricity: $e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac{5}{9}}=\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}$.
8. This matches option (b).