Equation of ellipse - class-XI

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).

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

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.

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.

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

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

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.

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.

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

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 $25{ x }^{ 2 }+16{ y }^{ 2 }=1600$

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