Let $P(x,y)$ be any point on the ellipse and PM be the perpendicular from P upon the directrix $3x+4y-5=0$.

Given foci $(\pm 5,0)$ and directrix $x=\cfrac{36}{5}$

Given $\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$ represents a ellipse

Comparing given equation with standard equation $r(1+e\cos\theta)=l$ where $l$ is semi latus rectum
Hence order is $D,B,C,A$

A perpendicular from a point on the conic to the axis is called an ordinate, and if produced to meet the conic again it is called a double ordinate. The double ordinate through the focus is called the $latus\ rectum$.

It is a fact & proof of it can be seen from higher education physics books

The sum of focal distances from a point of ellipse is 2 times the major axis.
For the given ellipse , length of semi major axis i.e. $b$ is $3$.
So required length $=2\times 3=6$
Option B is true

Equation of ellipse in standard form is 

equation of ellipse is ${ 9x }^{ 2 }+{ 4y }^{ 2 }-18x-8y-23=0$

The foci of the ellipse are also the foci of an hyperbola,
then we have, for the ellipse,

Given that focii are $(0,1), (0,-1)$

We know that equation of ellipse is

  $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $       …….(1)

Given that

  $ e=\sqrt{\dfrac{2}{5}} $

 $ \sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\dfrac{2}{5}} $

  $ 5{{a}^{2}}-5{{b}^{2}}=2{{a}^{2}} $

 $ {{a}^{2}}=\dfrac{5{{b}^{2}}}{3} $    …….(2)

$\because $ ellipse pass through (-3,1)

Then $x=-3, y=1$

Put in equation (1) we get

  $ \dfrac{{{\left( -3 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{1}^{2}}}{{{b}^{2}}}=1 $

 $ {{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}} $

 $ \dfrac{5{{b}^{2}}}{3}+9{{b}^{2}}=\dfrac{5{{b}^{2}}}{3}.{{b}^{2}} $

 $ {{b}^{2}}=\dfrac{32}{5} $    (From equation (1) and (2)  )

Put in equation (2) , we get ${{a}^{2}}=\dfrac{32}{3}$

the value of a and b put in equation (1), we get

  $ \dfrac{{{x}^{2}}}{\dfrac{32}{3}}+\dfrac{{{y}^{2}}}{\dfrac{32}{5}}=1 $

 $ 3{{x}^{2}}+5{{y}^{2}}=32 $

This is required equation

The length of latus Rectum of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

$\cfrac{x^2}{a^2+1}+\cfrac{y^2}{a^2+2}=1$

$\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1$

Given $x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$

Let the equation of the required ellipse be 

Given,ellipse equation as $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Length of major axis, $a=5$ and length of minor axis, $b=3$
$P$ is a point on the ellipse whose eccentricity is $\dfrac{\pi}{8}.$
We know that, sum of the distances of any point on the ellipse from its foci equal to twice the major axis.
Let $S,S'$ be foci of ellipse and $a,b$ as the length of major,minor axis respectively.
$\Rightarrow SP+S'P=2a$
$\Rightarrow SP+S'P=2 \times 5=10$

$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$ 
Here the coefficient of $\displaystyle y^{2}$ is + ive and that of $\displaystyle x^{2}$ is -ive and hence it represents a hyperbola whose transerse axis is vertical, i.e.
$\displaystyle a^{2}=4, b^{2}=9.$
$\displaystyle b^{2}= a^{2}\left ( e^{2}-1 \right )$
or $\displaystyle \frac{9}{4}+1=e^{2}\therefore e= \frac{\sqrt{13}}{2}$ 
Foci lie on y-axis $\displaystyle \left ( 0, \pm ae \right )$ i.e $\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$ 
$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$

The equation of the ellipse can be written as $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$

The length of latus rectum$(l)=4 \times SC$
where S is the focus and C is the vertex
Therefore, $l=4$

Consider Equation, $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$ to represent an ellipse equation.
$a>0,b>0,a\neq b$
Given,equation $\displaystyle\frac{x^2}{(8-t)}+\displaystyle\frac{y^2}{(t-4)}=1$
$\Rightarrow (8-t)>0\;$ and $\;(t-4)>0,(8-t)\neq(t-4)$
$\Rightarrow t\in(-\infty,8) \cap (4,\infty) \cap$ {$t\neq6$}
$\Rightarrow t\in(4,8)-${$6$}

$(3,5)$ lies on $25x^2+9y^2=450$

Given ellipse is $25{ x }^{ 2 }+16{ y }^{ 2 }=1600$

Here ${ \left( x-1 \right)  }^{ 2 }+{ \left( y-3 \right)  }^{ 2 }={ \left{ \cfrac { 5x-12y+17 }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right)  }^{ 2 } }  }  \right}  }^{ 2 }$

The vertices and foci of an ellipse are $\displaystyle \left( \pm 5,0 \right) $ and $\displaystyle \left( \pm 4,0 \right) $ respectively.
$\displaystyle \therefore \quad a=5$ and $\displaystyle ae=4$
$\displaystyle \Rightarrow \quad e=\frac { 4 }{ 5 } $
We know that,
$\displaystyle e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$\displaystyle \Rightarrow \quad \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 25 } \Rightarrow { b }^{ 2 }=9$
Hence, equation of an ellipse is
$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$

We know, if P is any point on the curve, then Sum of focal distances $=$ length of major axis
i.e., $SP + S'P = 2a$
$= 2(5) [\because a^{2} = 5^{2}]$
$= 10$

Given, ${x}^{2}+\dfrac{{y}^{2}}{4}=1$
It is in the form of  $ \dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$
Therefore, it represents an ellipse.

Given, $4{y}^{2}+{x}^{2}=25$

The differential equation is

The distance between the foci is $6$, so $c = 3$.
The sum of the distance from $(4, 2)$ to each of the foci is the major axis length,
so
$2a = \sqrt {(4 - 2)^{2} + (2 - 4)^{2}} + \sqrt {(4 - 2)^{2} + (2 + 2)^{2}}$
$= \sqrt {4 + 4} + \sqrt {4 + 16} = \sqrt {8} + \sqrt {20}$
$= 2\sqrt {2} + 2\sqrt {5} \Rightarrow a = \sqrt {2} + \sqrt {5}$
Also, for an ellipse,
$b^{2} = a^{2} - c^{2} = (\sqrt {2} + \sqrt {5})^{2} - 3^{2}$
$= 7 + 2\sqrt {10} = -2 + 2\sqrt {10}$.
Thus, we have
$(ab)^{2} = (7 + 2\sqrt {10})(-2 + 2\sqrt {10})$
$= -14 + 14\sqrt {10} - 4\sqrt {10} + 40$
$= 26 + 10\sqrt {10}$.

(A) Midpoint of the line segment joining the foci is called the centre of ellipse: TRUE

Given:

$ 2x^{2} + 3y - 8x - 18y + 35 = \lambda  $

$ 2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda  $

$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = \lambda  $

For $ \lambda = 0 $, then

$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = 0  $

Thus, the point is $ \left ( 2,3 \right ) $.

Hence, the correct option is ‘d’.

Given $x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$

$\begin{array}{l} { { Re } }\, \, \left( { \frac { 1 }{ z }  } \right) =c \ { { Re } }\, \, \left( { \frac { 1 }{ { x+iy } }  } \right) =c \ { { Re } }\, \, \left( { \frac { { x-iy } }{ { { x^{ 2 } }+{ y^{ 2 } } } }  } \right) =c \ \frac { x }{ { { x^{ 2 } }+{ y^{ 2 } } } } =c \ c\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) -{ x }=0 . \end{array}$

$9x^2+5(y^2-6y)=0$
$9x^2+5(y^2-6y+9)=45$
$9x^2+5(y-3)^2=45$
$\dfrac { x^2 }{ 5 }+\dfrac { (y-3)^2 }{ 9 }=1$  
$a^2 < b^2$
$a^2=b^2(1-e^2)$
$5=9(1-e^2)$
$\dfrac { 5 }{ 9 }=1-e^2$
$e^2=\dfrac{4}{9}$ 
$e=\dfrac{2}{3}$

We know that,

The equation of ellipse whose centre $(h, k)$.

$\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$

Given centre of ellipse $\left( h,k \right)=\left( 1,2 \right)$

Then, equation of ellipse $\dfrac{{{\left( x-1 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-2 \right)}^{2}}}{{{b}^{2}}}=1$          ……. (1)

But, the ellipse passes through given point $\left( x,y \right)=\left( 4,6 \right)$

By equation (1), we get

$ \dfrac{{{\left( 4-1 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( 6-2 \right)}^{2}}}{{{b}^{2}}}=1 $

$ \dfrac{{{3}^{2}}}{{{a}^{2}}}+\dfrac{{{4}^{2}}}{{{b}^{2}}}=1 $

$ \Rightarrow 9{{b}^{2}}+16{{a}^{2}}={{a}^{2}}{{b}^{2}} $

$\Rightarrow 16{{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}}$        …… (2)

Now, distance between focus and centre is $c=\sqrt{{{a}^{2}}-{{b}^{2}}}$

So,

$ c=\sqrt{{{\left( 1-6 \right)}^{2}}+{{\left( 2-2 \right)}^{2}}} $

$ c=\sqrt{{{5}^{2}}} $

$ c=5 $

$\sqrt{{{a}^{2}}-{{b}^{2}}}=5$

On squaring both sides, we get,

${{a}^{2}}-{{b}^{2}}=25$          …… (3)

By equation (2) and (3), we get

$ 9{{b}^{2}}+400+16{{b}^{2}}=25{{b}^{2}}+{{b}^{4}} $

$ 25{{b}^{2}}+400=25{{b}^{2}}+{{b}^{4}} $

$ 400={{b}^{4}} $

$ {{b}^{4}}-400=0 $

$ \left( {{b}^{2}}-20 \right)\left( {{b}^{2}}+20 \right)=0 $

${{b}^{2}}-20=0\,$            and        ${{b}^{2}}+20=0\,$

${{b}^{2}}=20$                and        ${{b}^{2}}=-20$      (Rejected)

Put the value of ${{b}^{2}}$ in equation (3) and we get,

${{a}^{2}}-{{b}^{2}}=25$

$ {{a}^{2}}-20=25 $

$ {{a}^{2}}=45 $

Now, put the value of ${{a}^{2}}$ and ${{b}^{2}}$ in equation (1), we get,

$\dfrac{{{\left( x-1 \right)}^{2}}}{45}+\dfrac{{{\left( y-2 \right)}^{2}}}{20}=1$

$3x^2+2y^2+6x-8y+5=0$
$\Rightarrow \dfrac {(x+1)^2}{2}+\dfrac {(y-2)^2}{3}=1$
Therefore, centre is $(-1, 2)$ and ellipse is vertical
$(\because b > a)$
$a^2=2, b^2=3$
Now $2=3(1-e^2)$
$\Rightarrow e=\dfrac {1}{\sqrt 3}$
Foci are $(-1, 2\pm be)$ and $(-1, 2\pm 1)$.
Hence, the foci are $(-1, 3)$ and $(-1, 1)$
The equations of the directrices are $y=2\pm \dfrac {b}{e}\Rightarrow y=5$ and $y=-1$

Given ellipse may be written as,
$\displaystyle9x^{2}+4\left ( y^{2}-\frac{15}{2}y+\frac{225}{16} \right

)=\frac{225}{4}$
or $\displaystyle \frac{x^{2}}{225/36}+\frac{\left (

y-15/4 \right )^{2}}{225/16}=1$
$\displaystyle \therefore

b^{2}=\frac{225}{16}, a^{2}=\frac{225}{36}$
$\displaystyle \therefore

e^{2}=1-\frac{a^{2}}{b^{2}}=1-\frac{16}{36}=1-\frac{4}{9}=\frac{5}{9}$
$\displaystyle

\therefore e=\frac{1}{3}\sqrt{5}.$

$\displaystyle 4x^{2}+y^{2}-8x+2y+1=0$
$\displaystyle 4(x^2-2x)+(y^2+2y)=-1$
$\displaystyle 4(x^2-2x+1)+(y^2+2y+1)=-1+5=4$
$\displaystyle 4(x-1)^2+(y+1)^2=4$
$\displaystyle \frac{\left ( x-1 \right )^{2}}{1}+\frac{\left (

y+1 \right )^{2}}{4}=1$
$\displaystyle \Rightarrow a^2 =1, b^2 = 4$ Clearly here $a^2< b^2$ so the axis of the ellipse is parallel to y-axis
Now,  $\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{3}}{2} \therefore $ Foci $\displaystyle \left

( 1, -1\pm \sqrt{3} \right ),$ Directrices  $\displaystyle y=-1\pm

\frac{4}{\sqrt{3}}.$

$4x^2+9y^2+8x+36y+4=0$
$\Rightarrow 4(x^2+2x)+9(y^2+4y)=-4$
$\Rightarrow 4(x^2+2x+1)+9(y^2+4y+4)=-4+4+36$
$\Rightarrow 4(x+1)^2+9(y+2)^2=36$
$\Rightarrow \dfrac{(x+1)^2}{9}+\dfrac{(y+2)^2}{4}=1$
$\therefore a^2=9,b^2=4$
Hence length of latus rectum is $=\dfrac{2b^2}{a}=\dfrac{8}{3}$

$12x^{ 2 }+4y^{ 2 }+24x-16y+25=0$

The equation of the ellipse can be written as $\dfrac{(x-3)^{2}}{4^{2}}+\dfrac{y^{2}}{5^{2}}=1$
$\therefore $ Minor axis is along the line $x-3=0$ and major axis is $y=0$ 
$(\because a^{2}=4^{2}< b^{2}=5^{2})$
$\therefore S\,$ and $ S^{'}$ are $(3,3),(3,-3)$
$\because 4^{2}=5^{2}(1-e^{2})\Rightarrow e=\dfrac{3}{5}\left ( \because ae=5\dfrac{3}{5}=3 \right )$ and $S\equiv(3,ae )$, $S^{'}\equiv(3,-ae).$
Ans: B

Given ${x}^{2}+{y}^{2}-2x+3y+2=0$