Ellipse Eccentricity Calculations - Class XII

The problem involves geometric properties of an ellipse. Solving for the orthocenter condition leads to the relation involving the eccentricity squared.

Given foci (3,1) and (1,1), the center is (2,1) and 2ae = 2, so ae = 1. Using the point (1,3), the sum of distances to the foci equals 2a.

Rewrite the ellipse equation in standard form by completing the square for x and y: 4(x + 1)^2 + 9(y + 2)^2 = 36, which simplifies to (x + 1)^2 / 9 + (y + 2)^2 / 4 = 1. Here a^2 = 9 and b^2 = 4, so the eccentricity e is sqrt(1 - b^2/a^2) = sqrt(1 - 4/9) = sqrt(5)/3.

For 5x^2 + 6xy + 5y^2 = 8, rotate the axes to eliminate the xy term. The eigenvalues of the matrix determine the semi-axes, leading to e = 1/sqrt(2).

If angle F1BF2 = 90 degrees, then in the right triangle F1OB, OB = OF1 = ae. Since b = ae, and b^2 = a^2(1-e^2), we get a^2e^2 = a^2(1-e^2), so 2e^2 = 1.

This is identical to 425790. The calculation leads to e = sqrt(2)-1.

The condition that the tangent meets the auxiliary circle at points subtending 90 degrees at the center relates the coordinates to the eccentricity.

Perimeter = PF1 + PF2 + F1F2 = 2a + 2ae = 15. Given 2a = 10, then 10 + 10e = 15, so 10e = 5, e = 1/2.

Using the normal equation at P(a cos theta, b sin theta) and the ratio of segments PQ and PR, one can solve for the eccentricity.

The normal at (ae, b^2/a) passes through (0, -2b). Using the normal equation (ax/cos theta - by/sin theta = a^2 - b^2) and substituting the point, we solve for e.

This is a duplicate of 425790 and 425795.

Roots of x^2 - 4x + 1 = 0 are 2 +/- sqrt(3). Thus a = 2 + sqrt(3) and b = 2 - sqrt(3). e^2 = 1 - b^2/a^2. Calculation shows e is between 1/3 and 1/2.

The eccentricity of an ellipse given the eccentric angles of the extremities of a focal chord is a standard derivation.

Vertices are (-4,1) and (6,1), so center is (1,1) and 2a = 10, a = 5. Focus lies on x - 2y = 2. With center (1,1), focus is (1+ae, 1). Plugging into x-2y=2: (1+ae) - 2(1) = 2 => ae = 3. e = 3/5.

The line 2ax + ysqrt(1-a^2) = 1 is of the form xcos(theta) + ysin(theta) = p where cos(theta) = 2a and sin(theta) = sqrt(1-a^2). Squaring and adding gives cos^2(theta) + sin^2(theta) = 4a^2 + 1 - a^2 = 3a^2 + 1 = 1, which is not quite right; however, the condition for a line to touch an ellipse x^2/A^2 + y^2/B^2 = 1 is p^2 = A^2cos^2(theta) + B^2*sin^2(theta). Comparing coefficients leads to A^2=1/4 and B^2=1, so e^2 = 1 - (1/4)/1 = 3/4, e = sqrt(3)/2.

The normal at (acos(theta), bsin(theta)) to the ellipse x^2/a^2 + y^2/b^2 = 1 is axsec(theta) - bycsc(theta) = a^2 - b^2. If this is a focal chord, it must pass through a focus (ae, 0). Substituting gives aaesec(theta) = a^2 - b^2. Thus a^2esec(theta) = a^2(1 - b^2/a^2) = a^2*e^2. Simplifying gives e = |sec(theta)|.

Given conic can be written as,

We have ${ e } _{ 1 }=\sqrt { 1-\cfrac { 16 }{ 25 }  } =\cfrac { 3 }{ 5 } $
$\because \quad { e } _{ 1 }{ e } _{ 2 }=1\Rightarrow { e } _{ 2 }=\cfrac { 5 }{ 3 } $
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of focii of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $. 
Let hyperbola is, $\cfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1..(1)$ 
given hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'A' is correct.

The equation represents the sum of distances from (x,y) to (-2,0) and (2,0) being 8. This is an ellipse with 2a = 8 (so a = 4) and 2ae = 4 (distance between foci (-2,0) and (2,0)). Thus ae = 2, so 4e = 2, e = 1/2.

$x^2 + 2y^2 - 2x + 3y + 2 = 0$
$\Rightarrow (x - 1)^2 + 2 (y + \dfrac34)^2 = \dfrac{1}{8}$
$\Rightarrow \dfrac {(x - 1)^2}{1 / 8} + \dfrac {(y + 3 / 4)^2}{1 / 16} = 1$
It is an ellipse with $a^2 = 1/8 , b^2 = 1/16$ .Hence its eccentricity
$e = \sqrt {1 - \dfrac{b^2}{a^2}} = \sqrt {1 - \dfrac8{16}} = \dfrac1{\sqrt 2}$

Given conic is $9x^2+5y^2-54x-40y+116=0$