Ellipse Questions in English

(B) The given equation is $|z - 2| + |z + 2| = 8$.
This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = 2$ and $z_2 = -2$.
Here,the distance between the fixed points (foci) is $|z_1 - z_2| = |2 - (-2)| = 4$.
Since the sum of distances from two fixed points is constant $(8)$ and this constant is greater than the distance between the fixed points $(4)$,the locus of $z$ is an ellipse.
Algebraically,let $z = x + iy$:
$\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 8$
$\sqrt{(x - 2)^2 + y^2} = 8 - \sqrt{(x + 2)^2 + y^2}$
Squaring both sides:
$(x - 2)^2 + y^2 = 64 + (x + 2)^2 + y^2 - 16\sqrt{(x + 2)^2 + y^2}$
$x^2 - 4x + 4 + y^2 = 64 + x^2 + 4x + 4 + y^2 - 16\sqrt{(x + 2)^2 + y^2}$
$-8x - 64 = -16\sqrt{(x + 2)^2 + y^2}$
$x + 8 = 2\sqrt{(x + 2)^2 + y^2}$
Squaring again:
$x^2 + 16x + 64 = 4(x^2 + 4x + 4 + y^2)$
$x^2 + 16x + 64 = 4x^2 + 16x + 16 + 4y^2$
$3x^2 + 4y^2 = 48$
$\frac{x^2}{16} + \frac{y^2}{12} = 1$,which is the equation of an ellipse.

(A) Let $P(x, y)$ be the moving point and $F_1(ae, 0)$,$F_2(-ae, 0)$ be the fixed points.
According to the definition of an ellipse,the sum of distances from two fixed points (foci) is constant $(2a)$.
$PF_1 + PF_2 = 2a$
$\sqrt{(x - ae)^2 + y^2} + \sqrt{(x + ae)^2 + y^2} = 2a$
$\sqrt{(x + ae)^2 + y^2} = 2a - \sqrt{(x - ae)^2 + y^2}$
Squaring both sides:
$(x + ae)^2 + y^2 = 4a^2 + (x - ae)^2 + y^2 - 4a\sqrt{(x - ae)^2 + y^2}$
$x^2 + 2aex + a^2e^2 + y^2 = 4a^2 + x^2 - 2aex + a^2e^2 + y^2 - 4a\sqrt{(x - ae)^2 + y^2}$
$4aex - 4a^2 = -4a\sqrt{(x - ae)^2 + y^2}$
$a - ex = \sqrt{(x - ae)^2 + y^2}$
Squaring again:
$a^2 - 2aex + e^2x^2 = x^2 - 2aex + a^2e^2 + y^2$
$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$
Dividing by $a^2(1 - e^2)$:
$\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1$.

(B) The definition of an ellipse is the locus of a point such that the sum of its distances from two fixed points (foci) is a constant,provided that the constant is greater than the distance between the two fixed points.
Given $PA + PB = 4$,where $4$ is the constant sum.
If the distance $AB < 4$,the locus of $P$ is an ellipse.
If $AB = 4$,the locus is the line segment $AB$.
If $AB > 4$,the locus is an empty set.
Assuming $AB < 4$,the locus is an ellipse.

(B) The equation of the line is $\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta - 1 = 0$,which can be written as $bx \cos \theta + ay \sin \theta - ab = 0$.
The perpendicular distance $p_1$ from $(\sqrt{a^2 - b^2}, 0)$ is $p_1 = \frac{|b(\sqrt{a^2 - b^2})\cos \theta + a(0)\sin \theta - ab|}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}} = \frac{|b\sqrt{a^2 - b^2}\cos \theta - ab|}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}}$.
The perpendicular distance $p_2$ from $(-\sqrt{a^2 - b^2}, 0)$ is $p_2 = \frac{|b(-\sqrt{a^2 - b^2})\cos \theta + a(0)\sin \theta - ab|}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}} = \frac{|-b\sqrt{a^2 - b^2}\cos \theta - ab|}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}}$.
The product $p_1 p_2 = \frac{|(b\sqrt{a^2 - b^2}\cos \theta - ab)(-b\sqrt{a^2 - b^2}\cos \theta - ab)|}{b^2 \cos^2 \theta + a^2 \sin^2 \theta}$.
Using $(x-y)(x+y) = x^2 - y^2$,the numerator becomes $|(ab)^2 - (b\sqrt{a^2 - b^2}\cos \theta)^2| = |a^2 b^2 - b^2(a^2 - b^2)\cos^2 \theta|$.
$p_1 p_2 = \frac{b^2 |a^2 - (a^2 - b^2)\cos^2 \theta|}{b^2 \cos^2 \theta + a^2 \sin^2 \theta} = \frac{b^2 |a^2 - a^2 \cos^2 \theta + b^2 \cos^2 \theta|}{b^2 \cos^2 \theta + a^2 \sin^2 \theta} = \frac{b^2 |a^2 \sin^2 \theta + b^2 \cos^2 \theta|}{b^2 \cos^2 \theta + a^2 \sin^2 \theta} = b^2$.

(D) Let $P(x, y)$ be a point in the plane. The given equation is $PF_1 + PF_2 = 4$,where $F_1 = (2, 0)$ and $F_2 = (-2, 0)$.
Here,the distance between the foci $F_1$ and $F_2$ is $d(F_1, F_2) = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$.
Since the sum of the distances from $P$ to two fixed points $F_1$ and $F_2$ is equal to the distance between the points themselves $(PF_1 + PF_2 = F_1F_2)$,the point $P$ must lie on the line segment connecting $F_1$ and $F_2$.
Thus,the equation represents the line segment on the $x$-axis between $x = -2$ and $x = 2$,which is a degenerate case of an ellipse where the minor axis is $0$.

(B) The formula for the length of the latus rectum of an ellipse is $\frac{2b^2}{a}$.
Given that the latus rectum is equal to half of the minor axis $(2b)$,we have:
$\frac{2b^2}{a} = \frac{1}{2} \times (2b)$
$\frac{2b^2}{a} = b$
Dividing both sides by $b$ (since $b \neq 0$):
$\frac{2b}{a} = 1 \implies \frac{b}{a} = \frac{1}{2}$
Squaring both sides:
$\frac{b^2}{a^2} = \frac{1}{4}$
The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the value of $\frac{b^2}{a^2}$:
$e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(C) For an ellipse with eccentricity $e$ and semi-major axis $a$,the distance between the directrices is given by $\frac{2a}{e}$.
The distance between the foci is given by $2ae$.
According to the given condition,the distance between the directrices is thrice the distance between the foci:
$\frac{2a}{e} = 3(2ae)$
$\frac{2a}{e} = 6ae$
$1 = 3e^2$
$e^2 = \frac{1}{3}$
$e = \frac{1}{\sqrt{3}}$.

(B) The standard equation of an ellipse with its centre at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through the points $(-3, 1)$ and $(2, -2)$,we substitute these coordinates into the equation:
For $(-3, 1)$: $\frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{1}{b^2} = 1$ (Equation $1$)
For $(2, -2)$: $\frac{2^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \implies \frac{4}{a^2} + \frac{4}{b^2} = 1 \implies \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ (Equation $2$)
Subtracting Equation $2$ from Equation $1$:
$(\frac{9}{a^2} + \frac{1}{b^2}) - (\frac{1}{a^2} + \frac{1}{b^2}) = 1 - \frac{1}{4}$
$\frac{8}{a^2} = \frac{3}{4} \implies a^2 = \frac{32}{3}$
Substituting $a^2$ into Equation $2$:
$\frac{1}{32/3} + \frac{1}{b^2} = \frac{1}{4} \implies \frac{3}{32} + \frac{1}{b^2} = \frac{8}{32}$
$\frac{1}{b^2} = \frac{5}{32} \implies b^2 = \frac{32}{5}$
Substituting $a^2$ and $b^2$ into the standard equation:
$\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1 \implies \frac{3x^2}{32} + \frac{5y^2}{32} = 1$
$3x^2 + 5y^2 = 32$.

(A) Given eccentricity $e = 5/8$ and distance between foci $2ae = 10$.
$2a(5/8) = 10 \implies a(5/4) = 10 \implies a = 8$.
Using the relation $b^2 = a^2(1 - e^2)$:
$b^2 = 8^2(1 - (5/8)^2) = 64(1 - 25/64) = 64 - 25 = 39$.
The length of the latus rectum is given by $\frac{2b^2}{a}$.
$\text{Latus rectum} = \frac{2 \times 39}{8} = \frac{39}{4}$.

(D) The standard equation of an ellipse with foci on the $x$-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Given foci are $(\pm ae, 0) = (\pm 1, 0)$,so $ae = 1$.
Given vertices are $(\pm a, 0) = (\pm 2, 0)$,so $a = 2$.
Using the relation $b^2 = a^2(1 - e^2) = a^2 - (ae)^2$,we get $b^2 = 2^2 - 1^2 = 4 - 1 = 3$.
Thus,$b = \sqrt{3}$.
The length of the minor axis is $2b = 2\sqrt{3}$.

(D) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.
Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The equations of the directrices for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $x = \pm \frac{a}{e}$.
Substituting the values,$x = \pm \frac{5}{3/5} = \pm \frac{25}{3}$,which implies $3x = \pm 25$.
Since $3x = \pm 25$ is not among the options $A, B, C$,the correct choice is $D$.

(B) Given eccentricity $e = 2/3$ and latus rectum $L = 2b^2/a = 5$.
Using the relation $e^2 = 1 - b^2/a^2$,we have $(2/3)^2 = 1 - b^2/a^2$,which implies $b^2/a^2 = 1 - 4/9 = 5/9$,so $b^2 = 5a^2/9$.
Substituting this into the latus rectum formula: $2(5a^2/9)/a = 5$.
This simplifies to $10a/9 = 5$,giving $a = 45/10 = 9/2$.
Then $b^2 = 5(9/2)^2/9 = 5(81/4)/9 = 45/4$.
Thus,$a^2 = 81/4$ and $b^2 = 45/4$.
The equation of the ellipse is $x^2/a^2 + y^2/b^2 = 1$,which becomes $x^2/(81/4) + y^2/(45/4) = 1$,or $\frac{4x^2}{81} + \frac{4y^2}{45} = 1$.

(A) Given the length of the latus rectum is $\frac{2b^2}{a} = 10$.
Given the minor axis is equal to the distance between the foci,so $2b = 2ae$,which implies $b = ae$.
We know that $b^2 = a^2(1 - e^2)$.
Substituting $b = ae$,we get $a^2e^2 = a^2(1 - e^2)$,which simplifies to $e^2 = 1 - e^2$,so $2e^2 = 1$ or $e = \frac{1}{\sqrt{2}}$.
Since $b = ae$,we have $b^2 = a^2e^2 = a^2(\frac{1}{2})$.
Substituting this into the latus rectum equation: $\frac{2(a^2/2)}{a} = 10$,which gives $a = 10$.
Then $b^2 = \frac{100}{2} = 50$,so $b = \sqrt{50} = 5\sqrt{2}$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{100} + \frac{y^2}{50} = 1$.
Multiplying by $100$,we get $x^2 + 2y^2 = 100$.

(C) Given the equation of the ellipse: $\frac{x^2}{36} + \frac{y^2}{20} = 1$.
Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we get $a^2 = 36$ and $b^2 = 20$,so $a = 6$ and $b = 2\sqrt{5}$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{20}{36}} = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3}$.
The equations of the directrices for an ellipse with $a > b$ are $x = \pm \frac{a}{e}$.
The distance between the directrices is $2 \times \frac{a}{e} = 2 \times \frac{6}{2/3} = 2 \times 9 = 18$.

(B) The given equation of the ellipse is $3x^2 + 4y^2 = 48$.
Dividing both sides by $48$,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 12$.
Thus,$a = 4$ and $b = \sqrt{12} = 2\sqrt{3}$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{12}{16}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
The distance between the foci is $2ae = 2 \times 4 \times \frac{1}{2} = 4$.

(A) The vertices of the ellipse are given as $(\pm a, 0) = (\pm 5, 0)$,so $a = 5$.
The foci are given as $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$.
Substituting $a = 5$,we get $5e = 4$,which means $e = \frac{4}{5}$.
For an ellipse,the relationship between $a, b,$ and $e$ is $b^2 = a^2(1 - e^2)$.
$b^2 = 5^2(1 - (\frac{4}{5})^2) = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.
Thus,$b = 3$.
The standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Multiplying by $225$,we obtain $9x^2 + 25y^2 = 225$.

(A) The foci are given by $(\pm ae, 0) = (\pm 5, 0)$,so $ae = 5$.
The equation of the directrix is $x = \frac{a}{e} = \frac{36}{5}$.
Multiplying the two equations: $(ae) \times (\frac{a}{e}) = 5 \times \frac{36}{5} \implies a^2 = 36 \implies a = 6$.
Substituting $a = 6$ into $ae = 5$,we get $6e = 5 \implies e = \frac{5}{6}$.
Using the relation $b^2 = a^2(1 - e^2)$,we have $b^2 = 36(1 - \frac{25}{36}) = 36(\frac{11}{36}) = 11$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{36} + \frac{y^2}{11} = 1$.

(D) Given eccentricity $e = \frac{1}{\sqrt{2}}$.
The formula for the latus rectum of an ellipse is $L = \frac{2b^2}{a}$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$.
Substituting the value of $e^2 = \frac{1}{2}$ into the equation:
$b^2 = a^2(1 - \frac{1}{2}) = a^2(\frac{1}{2}) = \frac{a^2}{2}$.
Now,substitute $b^2$ into the latus rectum formula:
$L = \frac{2(\frac{a^2}{2})}{a} = \frac{a^2}{a} = a$.
Since $a$ represents the semi-major axis,the latus rectum is equal to its semi-major axis.

(D) Given the equation of the ellipse is $5x^2 + 9y^2 = 45$.
Dividing both sides by $45$,we get $\frac{x^2}{9} + \frac{y^2}{5} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 5$,which implies $a = 3$ and $b = \sqrt{5}$.
The length of the latus rectum of an ellipse is given by the formula $\frac{2b^2}{a}$.
Substituting the values,we get $\text{Latus rectum} = \frac{2 \times 5}{3} = \frac{10}{3}$.

(D) The distance between a focus and the corresponding directrix of an ellipse is given by $\frac{a}{e} - ae = 8$.
Given eccentricity $e = \frac{1}{2}$.
Substituting $e$ in the equation: $\frac{a}{1/2} - a(\frac{1}{2}) = 8 \implies 2a - \frac{a}{2} = 8 \implies \frac{3a}{2} = 8 \implies a = \frac{16}{3}$.
The semi-minor axis $b$ is given by $b = a\sqrt{1 - e^2}$.
$b = \frac{16}{3}\sqrt{1 - (\frac{1}{2})^2} = \frac{16}{3}\sqrt{\frac{3}{4}} = \frac{16}{3} \times \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{3}$.
The length of the minor axis is $2b = 2 \times \frac{8\sqrt{3}}{3} = \frac{16\sqrt{3}}{3}$.
Since $\frac{16\sqrt{3}}{3}$ is not among the given options,the correct choice is $D$.

(C) Given the equation of the conic: $16x^2 + 7y^2 = 112$.
Dividing both sides by $112$,we get:
$\frac{16x^2}{112} + \frac{7y^2}{112} = 1$
$\frac{x^2}{7} + \frac{y^2}{16} = 1$.
This is an equation of an ellipse of the form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,where $a^2 = 16$ and $b^2 = 7$.
Since $a^2 > b^2$,the eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the values: $e = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{16 - 7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.

(B) The foci of an ellipse are given by $(\pm ae, 0)$.
The distance between the foci is $2ae$.
The length of the minor axis is $2b$.
According to the given condition,$2ae = 2b$,which implies $ae = b$.
We know the relation for an ellipse: $b^2 = a^2(1 - e^2)$.
Substituting $b = ae$ into the equation,we get $(ae)^2 = a^2(1 - e^2)$.
$a^2e^2 = a^2(1 - e^2)$.
Dividing both sides by $a^2$ (where $a \neq 0$),we get $e^2 = 1 - e^2$.
$2e^2 = 1$,which gives $e^2 = 1/2$.
Therefore,$e = 1/\sqrt{2}$.

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since it passes through $(-3, 1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$.
Multiplying by $a^2$,we get $9 + \frac{a^2}{b^2} = a^2$.....$(i)$
Given the eccentricity $e = \sqrt{\frac{2}{5}}$,we know $e^2 = 1 - \frac{b^2}{a^2}$.
So,$\frac{2}{5} = 1 - \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = \frac{3}{5}$,which means $\frac{a^2}{b^2} = \frac{5}{3}$.....$(ii)$
Substituting $(ii)$ into $(i)$,we get $9 + \frac{5}{3} = a^2 \Rightarrow a^2 = \frac{27 + 5}{3} = \frac{32}{3}$.
From $(ii)$,$b^2 = \frac{3}{5} a^2 = \frac{3}{5} \times \frac{32}{3} = \frac{32}{5}$.
Substituting $a^2$ and $b^2$ into the standard equation: $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$.
This simplifies to $\frac{3x^2}{32} + \frac{5y^2}{32} = 1$,or $3x^2 + 5y^2 = 32$.

(B) Given that the length of the major axis is $2b = 10$,so $b = 5$.
Given that the length of the minor axis is $2a = 8$,so $a = 4$.
Since the major axis lies along the $y$-axis,the standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{4^2} + \frac{y^2}{5^2} = 1$.
Thus,the equation is $\frac{x^2}{16} + \frac{y^2}{25} = 1$.

(A) Given: Centre $(h, k) = (0, 0)$,Focus $(0, 3)$,and semi-major axis $a = 5$.
Since the focus lies on the $y$-axis,the ellipse is vertical with the equation $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.
Here,the distance from the centre to the focus is $ae = 3$.
Given $a = 5$,we have $5e = 3$,so $e = \frac{3}{5}$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 25(1 - (\frac{3}{5})^2) = 25(1 - \frac{9}{25}) = 25 - 9 = 16$.
Thus,$b^2 = 16$ and $a^2 = 25$.
The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{25} = 1$.

(B) Given the vertex is $(0, 7)$ and the directrix is $y = 12$. Since the vertex lies on the $y$-axis,the major axis is along the $y$-axis.
For an ellipse with the major axis along the $y$-axis,the vertex is $(0, b)$ and the directrix is $y = b/e$.
Here,$b = 7$ and $b/e = 12$.
Thus,$e = b/12 = 7/12$.
The relationship between $a, b,$ and $e$ is $a^2 = b^2(1 - e^2)$.
$a^2 = 7^2(1 - (7/12)^2) = 49(1 - 49/144) = 49(95/144) = 4655/144$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values: $\frac{x^2}{4655/144} + \frac{y^2}{49} = 1$.
$\frac{144x^2}{4655} + \frac{y^2}{49} = 1$.
Multiplying by $4655$: $144x^2 + 95y^2 = 4655$.

(B) Given equation: $2x^2 + 3y^2 = 30$
Divide both sides by $30$:
$\frac{2x^2}{30} + \frac{3y^2}{30} = \frac{30}{30}$
$\frac{x^2}{15} + \frac{y^2}{10} = 1$
This is in the standard form of an ellipse,$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 15$ and $b^2 = 10$.
Therefore,the equation represents an ellipse.

(C) The length of the latus rectum of an ellipse is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.
Given the eccentricity $e = \frac{1}{\sqrt{2}}$,we use the relation $b^2 = a^2(1 - e^2)$.
Substituting $e^2 = \frac{1}{2}$,we get $b^2 = a^2(1 - \frac{1}{2}) = \frac{a^2}{2}$,so $a^2 = 2b^2$.
Substituting $b^2 = 4a$ into $a^2 = 2b^2$,we get $a^2 = 2(4a) = 8a$.
Since $a > 0$,$a = 8$. Then $a^2 = 64$.
Now,$b^2 = 4a = 4(8) = 32$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{64} + \frac{y^2}{32} = 1$.

(B) For an ellipse with equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$,the length of the latus rectum is $\frac{2b^2}{a}$ and the distance between the foci is $2ae$.
Given that the latus rectum is equal to the distance between the foci:
$\frac{2b^2}{a} = 2ae$
$\frac{b^2}{a} = ae$
$b^2 = a^2e$
Since $b^2 = a^2(1 - e^2)$,we substitute this into the equation:
$a^2(1 - e^2) = a^2e$
$1 - e^2 = e$
$e^2 + e - 1 = 0$
Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$
Since the eccentricity $e$ must be positive and $e < 1$,we take the positive root:
$e = \frac{\sqrt{5} - 1}{2}$

(B) Given the equation of the ellipse is $3x^2 + 4y^2 = 12$.
Dividing both sides by $12$,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = 3$.
Thus,$a = 2$ and $b = \sqrt{3}$.
The length of the latus rectum is given by the formula $\frac{2b^2}{a}$.
Substituting the values,we get $\frac{2 \times 3}{2} = 3$.

(A) The given equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{28} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 64$ and $b^2 = 28$.
Since $a^2 > b^2$,the eccentricity $e$ is given by the formula $e^2 = 1 - \frac{b^2}{a^2}$.
Substituting the values,we get $e^2 = 1 - \frac{28}{64}$.
$e^2 = \frac{64 - 28}{64} = \frac{36}{64}$.
Taking the square root,$e = \sqrt{\frac{36}{64}} = \frac{6}{8} = \frac{3}{4}$.

(D) Let the length of the major axis be $2a$ and the length of the minor axis be $2b$.
Given that the length of the major axis is three times the length of the minor axis:
$2a = 3(2b) \implies a = 3b$.
We know the formula for eccentricity $e$ of an ellipse is $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting $a = 3b$ into the formula:
$e = \sqrt{1 - \frac{b^2}{(3b)^2}}$
$e = \sqrt{1 - \frac{b^2}{9b^2}}$
$e = \sqrt{1 - \frac{1}{9}}$
$e = \sqrt{\frac{8}{9}}$
$e = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.
Thus,the eccentricity is $\frac{2\sqrt{2}}{3}$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The length of the latus rectum is $\frac{2b^2}{a}$ and the length of the major axis is $2a$.
Given that the length of the latus rectum is $\frac{1}{3}$ of the major axis:
$\frac{2b^2}{a} = \frac{1}{3} (2a)$
$\frac{b^2}{a} = \frac{a}{3}$
$3b^2 = a^2$
We know that for an ellipse,$b^2 = a^2(1 - e^2)$,where $e$ is the eccentricity.
Substituting $b^2$ in the equation:
$3a^2(1 - e^2) = a^2$
$3(1 - e^2) = 1$
$1 - e^2 = \frac{1}{3}$
$e^2 = 1 - \frac{1}{3} = \frac{2}{3}$
$e = \sqrt{\frac{2}{3}}$.

(D) The length of the string used to draw an ellipse with pins at the foci is equal to the sum of the major axis and the distance between the foci.
Given the major axis $2a = 6 \ cm$,so $a = 3 \ cm$.
Given the minor axis $2b = 4 \ cm$,so $b = 2 \ cm$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The distance between the pins (foci) is $2ae = 2 \times 3 \times \frac{\sqrt{5}}{3} = 2\sqrt{5} \ cm$.
The length of the string is $2a + 2ae = 6 + 2\sqrt{5} \ cm$.

(B) The given equation is $\frac{x^2}{2 - r} + \frac{y^2}{r - 5} + 1 = 0$.
Rearranging the terms,we get $\frac{x^2}{r - 2} + \frac{y^2}{5 - r} = 1$.
For this to represent an ellipse,the denominators must be positive,i.e.,$r - 2 > 0$ and $5 - r > 0$.
This implies $r > 2$ and $r < 5$.
Combining these,we get $2 < r < 5$.

(C) The locus of the point of intersection of perpendicular tangents to an ellipse is known as the director circle.
For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the equation of the director circle is given by $x^2 + y^2 = a^2 + b^2$.
Alternatively,let the point of intersection be $(h, k)$. The pair of tangents from $(h, k)$ to the ellipse is given by $SS_1 = T^2$,where $S = \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1$,$S_1 = \frac{h^2}{a^2} + \frac{k^2}{b^2} - 1$,and $T = \frac{hx}{a^2} + \frac{ky}{b^2} - 1$.
Expanding this,the condition for perpendicular tangents is that the sum of the coefficients of $x^2$ and $y^2$ must be zero.
This leads to $\frac{1}{a^2} (\frac{h^2}{a^2} + \frac{k^2}{b^2} - 1) - \frac{h^2}{a^4} + \frac{1}{b^2} (\frac{h^2}{a^2} + \frac{k^2}{b^2} - 1) - \frac{k^2}{b^4} = 0$,which simplifies to $h^2 + k^2 = a^2 + b^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = a^2 + b^2$.

(B) Given the equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{49} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 36$ and $b^2 = 49$.
Since $b^2 > a^2$,the major axis is along the $y$-axis,so $b = 7$ and $a = 6$.
The length of the latus rectum for an ellipse with the major axis along the $y$-axis is given by $\frac{2a^2}{b}$.
Substituting the values,we get $\text{Length} = \frac{2 \times 36}{7} = \frac{72}{7}$.

(C) The focal distance of any point $P(x, y)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $SP = a + ex$,where $S$ is the focus $(ae, 0)$.
For a point on the ellipse,the coordinates are given by $x = a \cos \theta$.
Substituting the value of $x$ into the formula,we get:
$SP = a + e(a \cos \theta)$
$SP = a(1 + e \cos \theta)$.

(B) Given the focus is at $(ae, 0) = (4, 0)$,we have $ae = 4$.
Given eccentricity $e = 4/5$,we find $a = 4 / (4/5) = 5$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 25(1 - 16/25) = 25(9/25) = 9$.
Thus,the equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$,which is $\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$.

(A) The given equation of the ellipse is $16x^2 + 25y^2 = 400$.
Dividing both sides by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci are given by $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.

(B) The given equation of the ellipse is $9x^2 + 25y^2 = 225$.
Dividing both sides by $225$,we get:
$\frac{9x^2}{225} + \frac{25y^2}{225} = 1$
$\frac{x^2}{25} + \frac{y^2}{9} = 1$
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 9$.
Since $a^2 > b^2$,the eccentricity $e$ is given by the formula:
$e = \sqrt{1 - \frac{b^2}{a^2}}$
$e = \sqrt{1 - \frac{9}{25}}$
$e = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

(C) The given equation is $25x^2 + 16y^2 = 100$.
Dividing both sides by $100$,we get $\frac{25x^2}{100} + \frac{16y^2}{100} = 1$,which simplifies to $\frac{x^2}{4} + \frac{y^2}{6.25} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = \frac{100}{16} = \frac{25}{4} = 6.25$.
Since $b^2 > a^2$,the ellipse is vertical,and the eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$.
Substituting the values: $4 = \frac{25}{4}(1 - e^2)$.
$1 - e^2 = \frac{16}{25}$.
$e^2 = 1 - \frac{16}{25} = \frac{9}{25}$.
Therefore,$e = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

(C) The given equation of the ellipse is $9x^2 + 4y^2 = 1$.
Dividing by $1$,we can rewrite it as $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1$.
Here,$a^2 = \frac{1}{9}$ and $b^2 = \frac{1}{4}$.
Since $b > a$,the major axis is along the $y$-axis.
The length of the latus rectum for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b > a$ is given by $\frac{2a^2}{b}$.
Substituting the values,we get $\text{Length} = \frac{2 \times (1/9)}{1/2} = \frac{2/9}{1/2} = \frac{4}{9}$.

(A) Let the variable point be $P(x, y)$.
According to the definition of a conic section,the distance from a fixed point (focus) is $e$ times the distance from a fixed line (directrix).
Here,the focus is $S(-2, 0)$,the directrix is $x = -\frac{9}{2}$,and the eccentricity $e = \frac{2}{3}$.
Since $e < 1$,the locus is an ellipse.
Mathematically,$\sqrt{(x + 2)^2 + y^2} = \frac{2}{3} |x + \frac{9}{2}|$.
Squaring both sides: $(x + 2)^2 + y^2 = \frac{4}{9} (x + \frac{9}{2})^2$.
$(x^2 + 4x + 4) + y^2 = \frac{4}{9} (x^2 + 9x + \frac{81}{4})$.
$x^2 + 4x + 4 + y^2 = \frac{4}{9}x^2 + 4x + 9$.
$\frac{5}{9}x^2 + y^2 = 5$.
Dividing by $5$: $\frac{x^2}{9} + \frac{y^2}{5} = 1$.
This is the equation of an ellipse.

(C) The given equation is $16x^2 + 25y^2 = 400$. Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$,which is an ellipse with $a^2 = 25$ and $b^2 = 16$.
Here,$a = 5$ and $b = 4$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
The foci are $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$,which match $F_1$ and $F_2$.
By the definition of an ellipse,the sum of the distances from any point $P$ on the ellipse to the two foci is equal to the length of the major axis,which is $2a$.
Thus,$PF_1 + PF_2 = 2a = 2 \times 5 = 10$.

(B) The given equation of the ellipse is $9x^2 + 36y^2 = 324$.
Dividing both sides by $324$,we get $\frac{9x^2}{324} + \frac{36y^2}{324} = 1$,which simplifies to $\frac{x^2}{36} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 36$,so $a = 6$.
By the definition of an ellipse,the sum of the focal distances of any point $P$ on the ellipse is equal to the length of the major axis,which is $2a$.
Therefore,$SP + S'P = 2a = 2 \times 6 = 12$.

(A) The foci are given by $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.
Given $e = \frac{1}{2}$,we have $a(\frac{1}{2}) = 2$,which implies $a = 4$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 16(1 - \frac{1}{4}) = 16(\frac{3}{4}) = 12$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.
Multiplying by $48$,we obtain $3x^2 + 4y^2 = 48$.

(C) Given the equation of the ellipse: $4x^2 + 9y^2 = 36$.
Dividing both sides by $36$,we get: $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
The formula for eccentricity $e$ is $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the values: $e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
Thus,the correct option is $C$.

(A) Given the equation of the ellipse: $25x^2 + 16y^2 = 400$.
Dividing both sides by $400$,we get: $\frac{x^2}{16} + \frac{y^2}{25} = 1$.
Here,$a^2 = 25$ and $b^2 = 16$. Since $a^2 > b^2$ is not the case here (as $25 > 16$),the major axis is along the $y$-axis.
For an ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the values: $e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.