Ellipse Questions in English

(D) Given,the distance between the foci of the ellipse is $2ae = 16$.
Given,the eccentricity of the ellipse is $e = \frac{1}{2}$.
We know that the length of the major axis of the ellipse is $2a$.
From the formula $2ae = 16$,we can write $2a = \frac{16}{e}$.
Substituting the value of $e = \frac{1}{2}$,we get $2a = \frac{16}{1/2} = 16 \times 2 = 32$.
Therefore,the length of the major axis is $32$.

(C) For the first ellipse $\frac{x^2}{169} + \frac{y^2}{25} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \frac{25}{169}}$.
For the second ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the eccentricity $e'$ is $e' = \sqrt{1 - \frac{b^2}{a^2}}$.
Given that the eccentricities are equal,$e = e'$,so $\sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{169}}$.
Squaring both sides,$1 - \frac{b^2}{a^2} = 1 - \frac{25}{169}$,which implies $\frac{b^2}{a^2} = \frac{25}{169}$.
Taking the square root,$\frac{b}{a} = \frac{5}{13}$.
Therefore,$\frac{a}{b} = \frac{13}{5}$.

(B) Given,the minor axis of the ellipse is $2b = 8$,so $b = 4$.
The eccentricity is $e = \frac{\sqrt{5}}{3}$.
We know the relation for an ellipse: $e^2 = 1 - \frac{b^2}{a^2}$.
Substituting the values: $\left(\frac{\sqrt{5}}{3}\right)^2 = 1 - \frac{4^2}{a^2}$.
$\frac{5}{9} = 1 - \frac{16}{a^2}$.
$\frac{16}{a^2} = 1 - \frac{5}{9} = \frac{4}{9}$.
$a^2 = \frac{16 \times 9}{4} = 36$,which gives $a = 6$.
The major axis of the ellipse is $2a = 2 \times 6 = 12$.

(D) The given equation is $9x^2 + 5y^2 = 45$.
Dividing both sides by $45$,we get $\frac{x^2}{5} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 5$ and $b^2 = 9$.
Since $b^2 > a^2$,the major axis is along the $y$-axis.
The eccentricity $e$ is given by $e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$.
Thus,$e = \frac{2}{3}$.
The distance between the foci is $2be = 2 \times \sqrt{9} \times \frac{2}{3} = 2 \times 3 \times \frac{2}{3} = 4$.

(B) Given that the eccentricity $e = \frac{1}{2}$ and the foci are at $(\pm ae, 0) = (\pm 1, 0)$.
From this,we have $ae = 1$.
Substituting $e = \frac{1}{2}$,we get $a(\frac{1}{2}) = 1$,which implies $a = 2$.
Using the relation $b^2 = a^2(1 - e^2)$,we calculate $b^2$:
$b^2 = 2^2(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
Since the foci are on the $x$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting $a^2 = 4$ and $b^2 = 3$,the equation is $\frac{x^2}{4} + \frac{y^2}{3} = 1$.

(A) By the definition of an ellipse,the sum of the focal distances of any point $P$ on the ellipse from the two foci $F_1$ and $F_2$ is constant.
This constant sum is equal to the length of the major axis of the ellipse.
Given the major axis length is $2a$,the sum of the focal distances is $2a$.

(D) Given that the distance between the foci is $2ae = 8$,so $ae = 4$.
The distance between the directrices is $\frac{2a}{e} = 18$,so $\frac{a}{e} = 9$.
Multiplying these two equations: $(ae) \times (\frac{a}{e}) = 4 \times 9$,which gives $a^2 = 36$,so $a = 6$.
Substituting $a = 6$ into $ae = 4$,we get $e = \frac{4}{6} = \frac{2}{3}$.
Using the relation $b^2 = a^2(1 - e^2)$,we have $b^2 = 36(1 - \frac{4}{9}) = 36(\frac{5}{9}) = 20$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.
Multiplying by $180$,we get $5x^2 + 9y^2 = 180$.

(B) Given equation: $4x^2 + 9y^2 - 16x - 54y + 61 = 0$
Rearranging the terms: $(4x^2 - 16x) + (9y^2 - 54y) = -61$
Factor out the coefficients: $4(x^2 - 4x) + 9(y^2 - 6y) = -61$
Complete the square: $4(x^2 - 4x + 4) + 9(y^2 - 6y + 9) = -61 + 16 + 81$
Simplify: $4(x - 2)^2 + 9(y - 3)^2 = 36$
Divide by $36$: $\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1$
Comparing this with the standard form $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$,the centre $(h, k)$ is $(2, 3).$

(A) Given equation: $4x^2 + 9y^2 - 8x - 36y + 4 = 0$
Rearranging terms: $4(x^2 - 2x) + 9(y^2 - 4y) = -4$
Completing the square: $4(x^2 - 2x + 1) + 9(y^2 - 4y + 4) = -4 + 4 + 36$
$4(x - 1)^2 + 9(y - 2)^2 = 36$
Dividing by $36$: $\frac{(x - 1)^2}{9} + \frac{(y - 2)^2}{4} = 1$
Here,$a^2 = 9$ and $b^2 = 4$,so $a = 3$ and $b = 2$.
The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$.

(B) Given equation: $4x^2 + y^2 - 8x + 2y + 1 = 0$
Rearranging the terms: $4(x^2 - 2x) + (y^2 + 2y) = -1$
Completing the square: $4(x^2 - 2x + 1) + (y^2 + 2y + 1) = -1 + 4 + 1$
$4(x - 1)^2 + (y + 1)^2 = 4$
Dividing by $4$: $\frac{(x - 1)^2}{1} + \frac{(y + 1)^2}{4} = 1$
Here,$a^2 = 4$ and $b^2 = 1$. Since $a^2 > b^2$,the ellipse is vertical.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(A) The length of the major axis is the distance between the vertices $(4, 0)$ and $(10, 0)$,which is $10 - 4 = 6$.
Thus,$2a = 6$,so $a = 3$.
The center of the ellipse is the midpoint of the vertices: $(\frac{4+10}{2}, 0) = (7, 0)$.
Given eccentricity $e = 1/2$,we find $b^2 = a^2(1 - e^2) = 3^2(1 - (1/2)^2) = 9(3/4) = 27/4$.
The equation of the ellipse is $\frac{(x - 7)^2}{a^2} + \frac{(y - 0)^2}{b^2} = 1$,which is $\frac{(x - 7)^2}{9} + \frac{y^2}{27/4} = 1$.
Multiplying by $27$,we get $3(x - 7)^2 + 4y^2 = 27$.
$3(x^2 - 14x + 49) + 4y^2 = 27$.
$3x^2 - 42x + 147 + 4y^2 = 27$.
$3x^2 + 4y^2 - 42x + 120 = 0$.

(B) Given the centre $(h, k) = (2, -3)$.
Since the $y$-coordinates of the centre,focus,and vertex are the same,the major axis is horizontal.
The distance from the centre to the focus is $ae = |3 - 2| = 1$.
The distance from the centre to the vertex is $a = |4 - 2| = 2$.
Thus,$e = \frac{ae}{a} = \frac{1}{2}$.
Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 2^2(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
The equation of the ellipse is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.
Substituting the values,we get $\frac{(x - 2)^2}{4} + \frac{(y + 3)^2}{3} = 1$.

(B) The given equation of the ellipse is $\frac{(x + y - 2)^2}{9} + \frac{(x - y)^2}{16} = 1$.
This is of the form $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$,where $X = x + y - 2$ and $Y = x - y$.
The centre of the ellipse is the point where $X = 0$ and $Y = 0$.
Setting $x + y - 2 = 0$ and $x - y = 0$,we get $x = y$.
Substituting $x = y$ into the first equation: $x + x - 2 = 0 \implies 2x = 2 \implies x = 1$.
Since $x = y$,we have $y = 1$.
Thus,the centre is $(1, 1)$.

(A) By the definition of an ellipse,the distance from any point $P(x, y)$ to the focus $S(-1, 1)$ is $e$ times the distance from $P$ to the directrix $L: x - y + 3 = 0$.
$SP^2 = e^2 \cdot d(P, L)^2$
$(x + 1)^2 + (y - 1)^2 = \left(\frac{1}{2}\right)^2 \cdot \frac{(x - y + 3)^2}{1^2 + (-1)^2}$
$(x^2 + 2x + 1 + y^2 - 2y + 1) = \frac{1}{4} \cdot \frac{(x - y + 3)^2}{2}$
$8(x^2 + y^2 + 2x - 2y + 2) = (x - y + 3)^2$
$8x^2 + 8y^2 + 16x - 16y + 16 = x^2 + y^2 + 9 - 2xy + 6x - 6y$
$7x^2 + 2xy + 7y^2 + 10x - 10y + 7 = 0$.

(A) The given equation is $25(x + 1)^2 + 9(y + 2)^2 = 225$.
Dividing by $225$,we get $\frac{(x + 1)^2}{9} + \frac{(y + 2)^2}{25} = 1$.
Here,$a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.
Since $a > b$,the major axis is vertical.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
The center of the ellipse is $(h, k) = (-1, -2)$.
The foci are given by $(h, k \pm ae) = (-1, -2 \pm 5 \times \frac{4}{5}) = (-1, -2 \pm 4)$.
Thus,the foci are $(-1, -2 + 4) = (-1, 2)$ and $(-1, -2 - 4) = (-1, -6)$.

(A) Given equations are $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$.
Dividing by the coefficients,we get $\frac{x}{3} = \cos t + \sin t$ and $\frac{y}{4} = \cos t - \sin t$.
Squaring both equations:
$(\frac{x}{3})^2 = \cos^2 t + \sin^2 t + 2 \sin t \cos t = 1 + \sin 2t$
$(\frac{y}{4})^2 = \cos^2 t + \sin^2 t - 2 \sin t \cos t = 1 - \sin 2t$
Adding the two squared equations:
$\frac{x^2}{9} + \frac{y^2}{16} = (1 + \sin 2t) + (1 - \sin 2t) = 2$
Dividing by $2$,we get $\frac{x^2}{18} + \frac{y^2}{32} = 1$.
This is the equation of an ellipse.

(C) Given the parametric equations $x = a \cos \theta$ and $y = b \sin \theta$.
Squaring and adding,we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2 \theta + \sin^2 \theta = 1$.
This is the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The eccentricity $e$ of an ellipse is given by the formula $e^2 = 1 - \frac{b^2}{a^2}$.
Simplifying this,we get $e^2 = \frac{a^2 - b^2}{a^2}$.

(D) Given equation: $4x^2 + 9y^2 + 8x + 36y + 4 = 0$
Rearranging the terms: $4(x^2 + 2x) + 9(y^2 + 4y) = -4$
Completing the square: $4(x^2 + 2x + 1) + 9(y^2 + 4y + 4) = -4 + 4 + 36$
$4(x + 1)^2 + 9(y + 2)^2 = 36$
Dividing by $36$: $\frac{(x + 1)^2}{9} + \frac{(y + 2)^2}{4} = 1$
Here,$a^2 = 9$ and $b^2 = 4$. Since $a^2 > b^2$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.
$e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.

(C) Given equation: $3x^2 + 4y^2 - 12x - 8y + 4 = 0$
Rearranging terms: $3(x^2 - 4x) + 4(y^2 - 2y) = -4$
Completing the square: $3(x - 2)^2 - 12 + 4(y - 1)^2 - 4 = -4$
$3(x - 2)^2 + 4(y - 1)^2 = 12$
Dividing by $12$: $\frac{(x - 2)^2}{4} + \frac{(y - 1)^2}{3} = 1$
Here,$a^2 = 4$ and $b^2 = 3$,so $a = 2$ and $b = \sqrt{3}$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
The foci in the shifted coordinate system $(X, Y)$ are $(\pm ae, 0)$,where $X = x - 2$ and $Y = y - 1$.
$ae = 2 \times \frac{1}{2} = 1$.
So,$X = \pm 1$ and $Y = 0$.
$x - 2 = 1 \implies x = 3$ and $y - 1 = 0 \implies y = 1$.
$x - 2 = -1 \implies x = 1$ and $y - 1 = 0 \implies y = 1$.
The foci are $(3, 1)$ and $(1, 1)$.

(C) The given equation is ${x^2} + 2{y^2} - 2x + 3y + 2 = 0$.
Completing the square for $x$ and $y$ terms:
$({x^2} - 2x + 1) + 2({y^2} + \frac{3}{2}y + \frac{9}{16}) = -2 + 1 + \frac{9}{8}$
${(x - 1)^2} + 2{(y + \frac{3}{4})^2} = \frac{1}{8}$
Dividing by $\frac{1}{8}$,we get:
$\frac{{{{(x - 1)}^2}}}{{1/8}} + \frac{{{{(y + 3/4)}^2}}}{{1/16}} = 1$
This is an ellipse of the form $\frac{{{{(x - h)}^2}}}{{{a^2}}} + \frac{{{{(y - k)}^2}}}{{{b^2}}} = 1$ where ${a^2} = \frac{1}{8}$ and ${b^2} = \frac{1}{{16}}$.
Since ${a^2} > {b^2}$,the eccentricity $e$ is given by ${b^2} = {a^2}(1 - {e^2})$.
$\frac{1}{{16}} = \frac{1}{8}(1 - {e^2})$
$1 - {e^2} = \frac{1}{2}$
${e^2} = \frac{1}{2}$
$e = \frac{1}{{\sqrt{2}}}$.

(C) The given equation of the ellipse is $25x^2 + 9y^2 - 150x - 90y + 225 = 0$.
Rearranging the terms,we get $25(x^2 - 6x) + 9(y^2 - 10y) = -225$.
Completing the square,$25(x - 3)^2 - 225 + 9(y - 5)^2 - 225 = -225$.
This simplifies to $25(x - 3)^2 + 9(y - 5)^2 = 225$.
Dividing by $225$,we get $\frac{(x - 3)^2}{9} + \frac{(y - 5)^2}{25} = 1$.
Here,$a^2 = 9$ and $b^2 = 25$. Since $b^2 > a^2$,the ellipse is vertical.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

(A) Given the equation of the ellipse: $\frac{(x - 1)^2}{9} + \frac{(y + 1)^2}{25} = 1$.
Comparing this with the standard form $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$,we have $b^2 = 9$ and $a^2 = 25$.
Since $a^2 > b^2$,the major axis is vertical.
The formula for eccentricity $e$ is $b^2 = a^2(1 - e^2)$.
Substituting the values: $9 = 25(1 - e^2)$.
$1 - e^2 = \frac{9}{25}$.
$e^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
$e = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

(A) Given the equation of the conic: $9x^2 + 4y^2 - 18x + 16y + 25 = 0$.
Rearranging the terms: $9(x^2 - 2x) + 4(y^2 + 4y) = -25$.
Completing the square: $9(x^2 - 2x + 1) + 4(y^2 + 4y + 4) = -25 + 9 + 16$.
$9(x - 1)^2 + 4(y + 2)^2 = 0$.
This represents a point ellipse (a degenerate conic) where the lengths of the axes are $0$. However,checking the standard form for an ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,if the equation was $9x^2 + 4y^2 - 18x + 16y + 25 = 1$,the lengths would be $2a$ and $2b$.
Given the original provided equation $9x^2 + 4y^2 - 6x + 4y + 1 = 0$:
$9(x^2 - \frac{2}{3}x + \frac{1}{9}) + 4(y^2 + y + \frac{1}{4}) = -1 + 1 + 1 = 1$.
$9(x - \frac{1}{3})^2 + 4(y + \frac{1}{2})^2 = 1$.
$\frac{(x - 1/3)^2}{1/9} + \frac{(y + 1/2)^2}{1/4} = 1$.
Here $a^2 = 1/9 \implies a = 1/3$ and $b^2 = 1/4 \implies b = 1/2$.
The lengths of the axes are $2a = 2/3$ and $2b = 1$.

(B) The given equation is $9x^2 + 5y^2 - 18x - 2y - 16 = 0$.
Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 0.4y) = 16$.
Completing the square,we have $9(x - 1)^2 - 9 + 5(y - 0.2)^2 - 0.2 = 16$,which simplifies to $9(x - 1)^2 + 5(y - 0.2)^2 = 25.2$.
Wait,let us re-evaluate the standard form: $9(x^2 - 2x + 1) + 5(y^2 - 0.4y + 0.04) = 16 + 9 + 0.2 = 25.2$.
Actually,the standard form is $\frac{(x-1)^2}{5/9} + \frac{(y-0.2)^2}{25.2/5} = 1$. Let us re-check the original equation coefficients.
Given $9x^2 + 5y^2 - 18x - 2y - 16 = 0 \implies 9(x-1)^2 - 9 + 5(y^2 - 0.4y + 0.04) - 0.2 = 16 \implies 9(x-1)^2 + 5(y-0.2)^2 = 25.2$.
Assuming the intended equation was $9x^2 + 5y^2 - 18x - 20y - 16 = 0$,then $9(x-1)^2 + 5(y-2)^2 = 16 + 9 + 20 = 45$.
Dividing by $45$,we get $\frac{(x-1)^2}{5} + \frac{(y-2)^2}{9} = 1$.
Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

(A) Given equation of the conic is $4x^2 + 16y^2 - 24x - 3y = 1$.
Rearranging the terms: $4(x^2 - 6x) + 16(y^2 - \frac{3}{16}y) = 1$.
Completing the square: $4(x^2 - 6x + 9) + 16(y^2 - \frac{3}{16}y + \frac{9}{1024}) = 1 + 36 + \frac{9}{64}$.
$4(x - 3)^2 + 16(y - \frac{3}{32})^2 = 37 + \frac{9}{64} = \frac{2377}{64}$.
Dividing by the constant: $\frac{(x - 3)^2}{2377/256} + \frac{(y - 3/32)^2}{2377/1024} = 1$.
Here,$a^2 = \frac{2377}{256}$ and $b^2 = \frac{2377}{1024}$.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2377/1024}{2377/256}} = \sqrt{1 - \frac{256}{1024}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) The equation of the ellipse is $\frac{x^2}{8} + \frac{y^2}{4} = 1$,where $a^2 = 8$ and $b^2 = 4$.
The line $y = mx + c$ is a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if $c^2 = a^2m^2 + b^2$.
Here,$m = 2$,$a^2 = 8$,and $b^2 = 4$.
Substituting these values,we get $c^2 = 8(2)^2 + 4$.
$c^2 = 8(4) + 4 = 32 + 4 = 36$.
Therefore,$c = \pm 6$.

(A) The equation of the ellipse is $S(x, y) = 2x^2 + 5y^2 - 20 = 0$.
To find the position of the point $(4, -3)$,we substitute the coordinates into the expression $S_1 = 2(4)^2 + 5(-3)^2 - 20$.
$S_1 = 2(16) + 5(9) - 20$.
$S_1 = 32 + 45 - 20 = 57$.
Since $S_1 > 0$,the point $(4, -3)$ lies outside the ellipse.

(C) The given ellipse is $x^2 + 16y^2 = 16$,which can be written as $\frac{x^2}{16} + \frac{y^2}{1} = 1$.
Here,$a^2 = 16$ and $b^2 = 1$.
The slope of the tangent $m = \tan(60^\circ) = \sqrt{3}$.
The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Substituting the values: $y = \sqrt{3}x \pm \sqrt{16(\sqrt{3})^2 + 1}$.
$y = \sqrt{3}x \pm \sqrt{16(3) + 1}$.
$y = \sqrt{3}x \pm \sqrt{48 + 1}$.
$y = \sqrt{3}x \pm \sqrt{49}$.
$y = \sqrt{3}x \pm 7$,which can be rewritten as $\sqrt{3}x - y \pm 7 = 0$.

(C) Given the equation of the ellipse: $4x^2 + 9y^2 - 16x - 54y + 61 = 0$.
Rewrite the equation by completing the square:
$4(x^2 - 4x) + 9(y^2 - 6y) + 61 = 0$
$4(x^2 - 4x + 4) + 9(y^2 - 6y + 9) = -61 + 16 + 81$
$4(x - 2)^2 + 9(y - 3)^2 = 36$
Divide by $36$:
$\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1$.
This is an ellipse with center $(2, 3)$ and major axis along the line $y = 3$.
Substitute the point $(1, 3)$ into the equation $f(x, y) = 4x^2 + 9y^2 - 16x - 54y + 61$:
$f(1, 3) = 4(1)^2 + 9(3)^2 - 16(1) - 54(3) + 61$
$f(1, 3) = 4 + 81 - 16 - 162 + 61 = -32$.
Since $f(1, 3) < 0$,the point lies inside the ellipse.
However,checking the major axis $y = 3$,the point $(1, 3)$ satisfies $y = 3$,so it lies on the major axis.

(A) The given line is $lx + my = n$,which can be rewritten as $y = -\frac{l}{m}x + \frac{n}{m}$.
Comparing this with the slope-intercept form of a tangent to an ellipse $y = mx_1 + c$,where $c^2 = a^2m_1^2 + b^2$,we have $m_1 = -\frac{l}{m}$ and $c = \frac{n}{m}$.
Substituting these into the condition $c^2 = a^2m_1^2 + b^2$,we get $\left(\frac{n}{m}\right)^2 = a^2\left(-\frac{l}{m}\right)^2 + b^2$.
Multiplying both sides by $m^2$,we obtain $n^2 = a^2l^2 + b^2m^2$.

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Any tangent to the ellipse is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$.
If this tangent passes through a point $(h, k)$,then $k - mh = \pm \sqrt{a^2m^2 + b^2}$.
Squaring both sides,we get $(k - mh)^2 = a^2m^2 + b^2$,which simplifies to $k^2 - 2mhk + m^2h^2 = a^2m^2 + b^2$.
Rearranging as a quadratic in $m$: $m^2(h^2 - a^2) - 2mhk + (k^2 - b^2) = 0$.
Since the tangents are mutually perpendicular,the product of their slopes $m_1m_2 = -1$.
For the quadratic equation $Am^2 + Bm + C = 0$,the product of roots is $\frac{C}{A}$.
Thus,$\frac{k^2 - b^2}{h^2 - a^2} = -1$.
$k^2 - b^2 = -(h^2 - a^2) \implies h^2 + k^2 = a^2 + b^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = a^2 + b^2$,which represents a circle known as the director circle.

(D) Given the ellipse equation is $\frac{x^2}{4} + \frac{y^2}{12} = 1$.
Check if the point $(1/4, 1/4)$ lies on the ellipse:
Substitute $x = 1/4$ and $y = 1/4$ into the equation:
$\frac{(1/4)^2}{4} + \frac{(1/4)^2}{12} = \frac{1/16}{4} + \frac{1/16}{12} = \frac{1}{64} + \frac{1}{192} = \frac{3+1}{192} = \frac{4}{192} = \frac{1}{48}$.
Since $\frac{1}{48} \neq 1$,the point $(1/4, 1/4)$ does not lie on the ellipse.
Therefore,the tangent at this point is not defined for this ellipse.
Thus,the correct option is $D$.

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Here,$a^2 = 16$ and $b^2 = 9$.
The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{16m^2 + 9}$.
Since the tangent passes through $(2, 3)$,we have $3 = 2m \pm \sqrt{16m^2 + 9}$.
Rearranging gives $3 - 2m = \pm \sqrt{16m^2 + 9}$.
Squaring both sides: $(3 - 2m)^2 = 16m^2 + 9$.
$9 - 12m + 4m^2 = 16m^2 + 9$.
$12m^2 + 12m = 0$.
$12m(m + 1) = 0$.
Thus,$m = 0$ or $m = -1$.
For $m = 0$,the tangent is $y - 3 = 0(x - 2) \implies y = 3$.
For $m = -1$,the tangent is $y - 3 = -1(x - 2) \implies y - 3 = -x + 2 \implies x + y = 5$.
Therefore,the equations of the tangents are $y = 3$ and $x + y = 5$.

(B) The equation of the tangent at point $(a \cos \theta, b \sin \theta)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
This can be rewritten in intercept form as $\frac{x}{(a / \cos \theta)} + \frac{y}{(b / \sin \theta)} = 1$.
Comparing this with the intercept form $\frac{x}{h} + \frac{y}{k} = 1$,we get $h = \frac{a}{\cos \theta}$ and $k = \frac{b}{\sin \theta}$.
Therefore,$\frac{a^2}{h^2} + \frac{b^2}{k^2} = \frac{a^2}{(a / \cos \theta)^2} + \frac{b^2}{(b / \sin \theta)^2} = \cos^2 \theta + \sin^2 \theta = 1$.

(A) The standard equation of an ellipse is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$,where the condition for the line $y = mx + c$ to be a tangent is $c^2 = A^2m^2 + B^2$.
In the given equation $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,we have $A^2 = b^2$ and $B^2 = a^2$.
Substituting these values into the condition $c^2 = A^2m^2 + B^2$,we get $c^2 = b^2m^2 + a^2$.
Therefore,$c = \pm \sqrt{b^2m^2 + a^2}$.

(C) The condition for the line $y = mx + c$ to intersect the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at real points is that the discriminant of the resulting quadratic equation must be non-negative,or equivalently,the line must not be strictly outside the ellipse.
The condition for tangency is $c^2 = a^2m^2 + b^2$.
For the line to intersect the ellipse at real points,the line must be a secant or tangent,which implies $c^2 \le a^2m^2 + b^2$.
Rearranging this inequality,we get $a^2m^2 \ge c^2 - b^2$.

(C) The locus of the point of intersection of two perpendicular tangents to an ellipse is known as the director circle.
For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the equation of the director circle is $x^2 + y^2 = a^2 + b^2$.
Given the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
Substituting these values into the formula,we get $x^2 + y^2 = 9 + 4 = 13$.
Thus,the locus is $x^2 + y^2 = 13$.

(C) The coordinates of any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentric angle $\theta$ are $(a \cos \theta, b \sin \theta)$.
The coordinates of the endpoints of the latus recta are $(ae, \pm \frac{b^2}{a})$.
Equating the coordinates,we have $a \cos \theta = ae$ and $b \sin \theta = \pm \frac{b^2}{a}$.
From $a \cos \theta = ae$,we get $\cos \theta = e$.
From $b \sin \theta = \pm \frac{b^2}{a}$,we get $\sin \theta = \pm \frac{b}{a}$.
Therefore,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\pm b/a}{e} = \pm \frac{b}{ae}$.
Thus,$\theta = \tan^{-1}\left( \pm \frac{b}{ae} \right)$.

(D) The given equation of the ellipse is $x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{6} + \frac{y^2}{2} = 1$.
Here,$a^2 = 6$ and $b^2 = 2$,so $a = \sqrt{6}$ and $b = \sqrt{2}$.
Any point on the ellipse can be represented as $(a \cos \theta, b \sin \theta) = (\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)$,where $\theta$ is the eccentric angle.
The distance of this point from the centre $(0, 0)$ is given as $2$ units.
So,$\sqrt{(\sqrt{6} \cos \theta)^2 + (\sqrt{2} \sin \theta)^2} = 2$.
Squaring both sides,we get $6 \cos^2 \theta + 2 \sin^2 \theta = 4$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we have $6 \cos^2 \theta + 2(1 - \cos^2 \theta) = 4$.
$4 \cos^2 \theta + 2 = 4 \implies 4 \cos^2 \theta = 2 \implies \cos^2 \theta = \frac{1}{2}$.
Thus,$\cos \theta = \pm \frac{1}{\sqrt{2}}$.
This gives $\theta = \frac{\pi}{4}$ or $\theta = \frac{3\pi}{4}$.

(C) The given equation is $9x^2 + 5y^2 - 30y = 0$.
Completing the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$.
$9x^2 + 5(y^2 - 6y + 9) = 45$.
Dividing by $45$,we get $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$.
Since the denominator of $(y - 3)^2$ is greater than the denominator of $x^2$ $(9 > 5)$,the major axis is parallel to the $y$-axis.
The vertices are found by setting $x = 0$ in the equation $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$,which gives $\frac{(y - 3)^2}{9} = 1$,so $(y - 3)^2 = 9$.
Thus,$y - 3 = \pm 3$,which means $y = 6$ or $y = 0$.
The tangents at the ends of the major axis are horizontal lines passing through these vertices,which are $y = 0$ and $y = 6$.

(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.
At the point $(a \cos \theta, b \sin \theta)$,the slope of the tangent is $m_t = -\frac{b^2 (a \cos \theta)}{a^2 (b \sin \theta)} = -\frac{b \cos \theta}{a \sin \theta}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = \frac{a \sin \theta}{b \cos \theta}$.
The equation of the normal is $y - b \sin \theta = \frac{a \sin \theta}{b \cos \theta} (x - a \cos \theta)$.
Multiplying by $b \cos \theta$,we get $by \cos \theta - b^2 \sin \theta \cos \theta = ax \sin \theta - a^2 \sin \theta \cos \theta$.
Rearranging terms,$ax \sin \theta - by \cos \theta = (a^2 - b^2) \sin \theta \cos \theta$.
Dividing by $\sin \theta \cos \theta$,we get $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$.

(B) The equation of the normal at point $P(a\cos \theta, b\sin \theta)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $ax \sec \theta - by \csc \theta = a^2 - b^2$.
Given $a^2 = 14$ and $b^2 = 5$,the normal equation is $\frac{14x}{a\cos \theta} - \frac{5y}{b\sin \theta} = 14 - 5 = 9$.
Substituting the coordinates of point $Q(a\cos 2\theta, b\sin 2\theta)$ into the normal equation:
$\frac{14(a\cos 2\theta)}{a\cos \theta} - \frac{5(b\sin 2\theta)}{b\sin \theta} = 9$
$14 \frac{\cos 2\theta}{\cos \theta} - 5 \frac{\sin 2\theta}{\sin \theta} = 9$
Using $\cos 2\theta = 2\cos^2 \theta - 1$ and $\sin 2\theta = 2\sin \theta \cos \theta$:
$14 \frac{2\cos^2 \theta - 1}{\cos \theta} - 5 \frac{2\sin \theta \cos \theta}{\sin \theta} = 9$
$28\cos \theta - \frac{14}{\cos \theta} - 10\cos \theta = 9$
$18\cos \theta - \frac{14}{\cos \theta} = 9$
$18\cos^2 \theta - 9\cos \theta - 14 = 0$
Factoring the quadratic equation:
$(6\cos \theta - 7)(3\cos \theta + 2) = 0$
Since $|\cos \theta| \leq 1$,we have $3\cos \theta + 2 = 0$,which gives $\cos \theta = -\frac{2}{3}$.

(C) The equation of the line is $mx - y + c = 0$.
Comparing this with $Lx + My + N = 0$,we have $L = m$,$M = -1$,and $N = c$.
The condition for the line $Lx + My + N = 0$ to be a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{a^2}{L^2} + \frac{b^2}{M^2} = \frac{(a^2 - b^2)^2}{N^2}$.
Substituting the values of $L, M, N$:
$\frac{a^2}{m^2} + \frac{b^2}{(-1)^2} = \frac{(a^2 - b^2)^2}{c^2}$
$\frac{a^2 + b^2m^2}{m^2} = \frac{(a^2 - b^2)^2}{c^2}$
$c^2 = \frac{m^2(a^2 - b^2)^2}{a^2 + b^2m^2}$
$c = \pm \frac{(a^2 - b^2)m}{\sqrt{a^2 + b^2m^2}}$.

(D) Given the ellipse equation: $9x^2 + 5y^2 = 45$.
Dividing by $45$,we get $\frac{x^2}{5} + \frac{y^2}{9} = 1$.
Here,$a^2 = 5$ and $b^2 = 9$.
The point $(x_1, y_1) = (0, 3)$ lies on the ellipse.
The equation of the normal at $(x_1, y_1)$ is given by $\frac{a^2(x - x_1)}{x_1} = \frac{b^2(y - y_1)}{y_1}$.
Substituting the values: $\frac{5(x - 0)}{0} = \frac{9(y - 3)}{3}$.
This implies $\frac{5x}{0} = 3(y - 3)$.
For the expression to be defined,$x$ must be $0$.
Thus,the equation of the normal is $x = 0$,which represents the $y$-axis.

(B) The equation of the normal at point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.
Given the ellipse $9x^2 + 16y^2 = 180$,we divide by $180$ to get $\frac{x^2}{20} + \frac{y^2}{11.25} = 1$.
Here,$a^2 = 20$ and $b^2 = \frac{180}{16} = 11.25$.
At point $(2, 3)$,the equation of the normal is $\frac{20x}{2} - \frac{11.25y}{3} = 20 - 11.25$.
$10x - 3.75y = 8.75$.
Multiplying by $4$ to clear decimals: $40x - 15y = 35$.
Dividing by $5$: $8x - 3y = 7$,which simplifies to $3y - 8x + 7 = 0$.

(D) The equation of any normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $\phi$ is given by $ax \sec \phi - by \csc \phi = a^2 - b^2$ ... $(i)$.
The given line is $x \cos \alpha + y \sin \alpha = p$.
If $(i)$ and the given line represent the same line,then the ratios of their coefficients must be equal:
$\frac{a \sec \phi}{\cos \alpha} = \frac{-b \csc \phi}{\sin \alpha} = \frac{a^2 - b^2}{p}$.
From this,we get:
$\cos \phi = \frac{ap}{(a^2 - b^2) \cos \alpha}$ and $\sin \phi = \frac{-bp}{(a^2 - b^2) \sin \alpha}$.
Using the identity $\sin^2 \phi + \cos^2 \phi = 1$,we substitute the values:
$\frac{b^2 p^2}{(a^2 - b^2)^2 \sin^2 \alpha} + \frac{a^2 p^2}{(a^2 - b^2)^2 \cos^2 \alpha} = 1$.
Multiplying by $(a^2 - b^2)^2$,we get:
$p^2 \left( \frac{b^2}{\sin^2 \alpha} + \frac{a^2}{\cos^2 \alpha} \right) = (a^2 - b^2)^2$.
Thus,$p^2(a^2 \sec^2 \alpha + b^2 \csc^2 \alpha) = (a^2 - b^2)^2$.

(B) The equation of any normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos \theta, b \sin \theta)$ is given by $ax \sec \theta - by \csc \theta = a^2 - b^2$ $(i)$.
The given line is $lx + my + n = 0$ $(ii)$.
If $(i)$ and $(ii)$ represent the same line,then the ratios of their coefficients must be equal:
$\frac{a \sec \theta}{l} = \frac{-b \csc \theta}{m} = \frac{-(a^2 - b^2)}{n}$.
From this,we get:
$\cos \theta = \frac{-an}{l(a^2 - b^2)}$ and $\sin \theta = \frac{bn}{m(a^2 - b^2)}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$\frac{a^2 n^2}{l^2(a^2 - b^2)^2} + \frac{b^2 n^2}{m^2(a^2 - b^2)^2} = 1$.
Multiplying both sides by $(a^2 - b^2)^2 / n^2$,we get:
$\frac{a^2}{l^2} + \frac{b^2}{m^2} = \frac{(a^2 - b^2)^2}{n^2}$.

(A) Given the equation of the ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
The equation of the tangent at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.
Substituting $(x_1, y_1) = (3, -2)$,$a^2 = 9$,and $b^2 = 4$:
$\frac{3x}{9} + \frac{-2y}{4} = 1 \Rightarrow \frac{x}{3} - \frac{y}{2} = 1$.
The slope of the tangent is $m_t = \frac{1/3}{1/2} = \frac{2}{3}$.
The slope of the normal $m_n$ is $-\frac{1}{m_t} = -\frac{3}{2}$.
The equation of the normal at $(3, -2)$ is $y - (-2) = -\frac{3}{2}(x - 3)$.
$2(y + 2) = -3(x - 3)$ $\Rightarrow 2y + 4 = -3x + 9$ $\Rightarrow 3x + 2y = 5$.
Dividing by $6$,we get $\frac{3x}{6} + \frac{2y}{6} = \frac{5}{6} \Rightarrow \frac{x}{2} + \frac{y}{3} = \frac{5}{6}$.

(A) The given conic is $x^2 + \frac{y^2}{4} = 1$,which is an ellipse with $a^2 = 1$ and $b^2 = 4$,so $a = 1$ and $b = 2$.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is given by $ax \sec \theta - by \csc \theta = a^2 - b^2$.
Substituting $a = 1$ and $b = 2$,the normal equation becomes $x \sec \theta - 2y \csc \theta = 1 - 4 = -3$.
Multiplying by $\cos \theta$,we get $x - 2y \cot \theta = -3 \cos \theta$.
The given line is $2x - \frac{8}{3}\lambda y = -3$. Rewriting this as $x - \frac{4}{3}\lambda y = -\frac{3}{2}$.
Comparing the two equations,we have $\cot \theta = \frac{2}{3}\lambda$ and $-3 \cos \theta = -\frac{3}{2}$,which implies $\cos \theta = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,then $\sin \theta = \frac{\sqrt{3}}{2}$ and $\cot \theta = \frac{1}{\sqrt{3}}$.
Equating $\frac{2}{3}\lambda = \frac{1}{\sqrt{3}}$ gives $\lambda = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Wait,re-evaluating the comparison: The normal equation is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$. For $a=1, b=2$,this is $\frac{x}{\cos \theta} - \frac{2y}{\sin \theta} = -3$.
Comparing $\frac{x}{\cos \theta} - \frac{2y}{\sin \theta} = -3$ with $2x - \frac{8}{3}\lambda y = -3$,we have $\frac{1}{\cos \theta} = \frac{2}{2} = 1 \implies \cos \theta = 1$ (not possible) or scaling the equation.
Scaling the normal equation by $2$: $\frac{2x}{\cos \theta} - \frac{4y}{\sin \theta} = -6$. This does not match.
Correct approach: $\frac{x}{\cos \theta} = \frac{2}{k}$ and $\frac{2y}{\sin \theta} = \frac{8\lambda y}{3k}$ and $-3 = -3k$. Thus $k=1$.
Then $\cos \theta = 1/2$ and $\sin \theta = 2 / (8\lambda/3) = 6/(8\lambda) = 3/(4\lambda)$.
Since $\cos^2 \theta + \sin^2 \theta = 1$,$(1/2)^2 + (3/4\lambda)^2 = 1 \implies 9/(16\lambda^2) = 3/4 \implies 16\lambda^2 = 12 \implies \lambda^2 = 3/4 \implies \lambda = \sqrt{3}/2$.

(A) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,two diameters $y = m_1x$ and $y = m_2x$ are conjugate if $m_1m_2 = -\frac{b^2}{a^2}$.
Given the diameter $y = \frac{b}{a}x$,we have $m_1 = \frac{b}{a}$.
Let the conjugate diameter be $y = m_2x$.
Then,$\frac{b}{a} \times m_2 = -\frac{b^2}{a^2}$.
Solving for $m_2$,we get $m_2 = -\frac{b^2}{a^2} \times \frac{a}{b} = -\frac{b}{a}$.
Thus,the equation of the conjugate diameter is $y = -\frac{b}{a}x$.