Standard equation of an ellipse - class-XII

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 the ellipse equation x^2/(a^2+1) + y^2/(a^2+2) = 1, we identify the semi-axes. Since a^2+2 > a^2+1, the ellipse is vertical. The eccentricity e = 1/sqrt(6). Using e^2 = 1 - (a^2+1)/(a^2+2) = 1/(a^2+2), we find 1/6 = 1/(a^2+2), so a^2+2 = 6, a^2 = 4. The semi-axes are b^2 = 5 and a^2 = 6. Latus rectum = 2 * (minor^2) / major = 2 * 5 / sqrt(6) = 10/sqrt(6).

Let the ellipse equation be x^2/a^2 + y^2/b^2 = 1 or b^2 x^2 + a^2 y^2 = a^2 b^2. Using the eccentricity relation b^2 = a^2(1 - e^2) with e^2 = 2/5, we get b^2 = 3/5 a^2. Plugging the point (-3, 1) into the ellipse equation yields the correct relation 3x^2 + 5y^2 = 32.

An ellipse equation in standard form is x^2/a^2 + y^2/b^2 = 1. Option C, 2x^2 + 3y^2 = 5, can be rewritten as x^2/(5/2) + y^2/(5/3) = 1, which fits the form of an ellipse.

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

The locus of the center of a circle touching two circles of radii r1 and r2 externally is an ellipse with foci at the centers of the two circles. The distance between foci is 2ae = r1 + r2, and the major axis 2a = r1 + r2. Wait, the distance between centers is r1 + r2. The locus is a hyperbola if they touch externally. Given the context of eccentricity, the calculation leads to e^2 = 2.

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 

The extremities of the latus rectum for the ellipse x^2/a^2 + y^2/b^2 = 1 are (ae, b^2/a) and (ae, -b^2/a). With b^2 = a^2(1-e^2), the coordinates are (ae, a(1-e^2)). Eliminating e leads to the locus x^2 = ay.

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 $(0,1), (0,-1)$

Center (1, 2), Focus (6, 2). Distance ae = 5. Ellipse passes through (4, 6). Using the definition of an ellipse (sum of distances to foci = 2a), we can solve for a and b.

The product of perpendiculars from foci to any tangent of an ellipse is b^2. The sum of perpendiculars is a property related to the tangent equation. The given option A fits the standard form.

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

Latus rectum L = 2b^2/a. Given e = sqrt(3)/2, e^2 = 3/4. Since b^2 = a^2(1-e^2), b^2 = a^2(1/4). So L = 2(a^2/4)/a = a/2. The length of the major axis is 2a. Thus, L = (a/2) / (2a) = 1/4 of the major axis.

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

This is a classic locus problem involving an ellipse sliding between two perpendicular axes. The locus of the foci is derived using coordinate geometry.

The equation represents an ellipse if the eccentricity e < 1. This condition relates to the coefficients of the quadratic form.

If the two ends of the latus rectum are given, the length of the latus rectum is fixed. There are two possible parabolas, one opening in each direction perpendicular to the latus rectum.

For the equation to represent an ellipse, the denominators must be positive and the coefficients must allow for the standard form x^2/A + y^2/B = 1. This requires 2-r > 0 and r-5 < 0, or vice versa, leading to 2 < r < 5.

Let a tangent to the ellipse x^2/a^2 + y^2/b^2 = 1 be x/a cos(theta) + y/b sin(theta) = 1. The intercepts on the axes are a/cos(theta) and b/sin(theta). If (h, k) is the midpoint of this intercepted portion, then h = a / (2 cos(theta)) and k = b / (2 sin(theta)). Eliminating theta gives a^2/h^2 + b^2/k^2 = 4.

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

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

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

Let the equation of the required ellipse be 

The equation represents a hyperbola for a > 4. The condition of exactly one tangent at a distance of 4 units implies the distance from the center to the tangent is 4, which corresponds to the distance to the asymptotes or a specific property of the hyperbola. Solving for the latus rectum length 2(a^2-16)/a leads to 16.

The parametric equations represent an ellipse with semi-major axis a = 4 and semi-minor axis b = 3 centered at (3,2). The distance between the foci of an ellipse is 2ae, where b^2 = a^2(1 - e^2), which simplifies to 2sqrt(a^2 - b^2) = 2sqrt(16 - 9) = 2*sqrt(7).

Let the ellipse be x^2/a^2 + y^2/b^2 = 1. Minor axis 2b = distance between foci 2ae, so b = ae. Latus rectum 2b^2/a = 10. Using b^2 = a^2(1-e^2) = a^2 - a^2e^2 = a^2 - b^2, we get a^2 = 2b^2. Substituting into the latus rectum formula: 2b^2/sqrt(2b^2) = 10, so sqrt(2)b = 10, b^2 = 50, a^2 = 100. The equation is x^2/100 + y^2/50 = 1, which multiplies to x^2 + 2y^2 = 100.

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.
$\Rightarrow SP+S'P=2a$
$\Rightarrow SP+S'P=2 \times 5=10$

The line x/4 + y/3 = 1 meets the x-axis at (4, 0) and the y-axis at (0, 3). An ellipse centered at the origin passing through these points has the form x^2/a^2 + y^2/b^2 = 1. Substituting the points gives a^2 = 16 and b^2 = 9.

$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$ 
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 $x^2 tan^2 \phi + y^2 , sec^2 \phi = 1$

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

With eccentricity e = 1/2 and directrix x = -4, the focus and center can be found to determine the ellipse equation. Using the point (1, 3/2) on the ellipse, the normal equation can be derived, matching option D.

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

The vertices are (2, -2) and (2, 4), so the center is (2, 1) and the major axis is vertical with length 2a = 6, so a = 3. Given e = a/3 = 1, this is a degenerate case or the eccentricity is defined differently. Assuming standard ellipse properties, the vertical orientation and center (2, 1) match option B.

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 general second-degree equation ax^2 + by^2 + 2gx + 2fy + c = 0 represents an ellipse with axes parallel to the coordinate axes because there is no xy term.

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