List of top Mathematics Questions on Ellipse

Let the length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\dfrac{3}{4}+2t-t^2$, then $(a^2+b^2)$ is equal to

Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$

Let the ellipse \[ E=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] have eccentricity equal to the greatest value of the function \[ f(t)=-\frac34+2t-t^2 \] and the length of its latus rectum is $30$. Find the value of $a^2+b^2$.
Let the ellipse \[ E=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] have eccentricity equal to the greatest value of the function \[ f(t)=-\frac34+2t-t^2 \] and the length of its latus rectum is $30$. Find the value of $a^2+b^2$.
Let the length of a latus rectum of an ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ be 10. If its eccentricity is $ e $, and the minimum value of the function $ f(t) = t^2 + t + \frac{11}{12} $, where $ t \in \mathbb{R} $, then $ a^2 + b^2 $ is equal to:
If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is:
Let $ e_1 $ and $ e_2 $ be the eccentricities of the ellipse $ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $ and the hyperbola $ \frac{x^2}{16} - \frac{y^2}{b^2} = 1, $ respectively. If $ b<5 $ and $ e_1 e_2 = 1 $, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:
Let $ A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\} $ and $ B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\}. $ Then:
If the equation of an ellipse \( E \) is \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \), then the length of the latus rectum of \( E \) is?
Let the ellipse \( E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a > b \) and \( E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \), \( A < B \), have the same eccentricity \( \sqrt{\frac{1}{3}} \). Let the product of their lengths of latus rectums be \( \sqrt{\frac{32}{3}} \) and the distance between the foci of \( E_1 \) be 4. If \( E_1 \) and \( E_2 \) meet at \( A \), \( B \), \( C \), and \( D \), then the area of the quadrilateral ABCD equals:
Let the product of the focal distances of the point $$ \left( \sqrt{3}, \frac{1}{2} \right) $$ on the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a > b), $$ be $ \frac{7}{4} $. Then the absolute difference of the eccentricities of two such ellipses is:
Length of latus rectum of ellipse \[ \frac{x^2}{9} + \frac{y^2}{16} = 1 \]
Find the center of the ellipse given by the equation \[ 4x^2 + 24x + 9y^2 - 18y + 9 = 0 \]
Let \( f(x) = x^2 + 9 \), \( g(x) = \frac{x}{x-9} \), and \[ a = f \circ g(10), \, b = g \circ f(3). \]
If \( e \) and \( l \) denote the eccentricity and the length of the latus rectum of the ellipse \[ \frac{x^2}{a} + \frac{y^2}{b} = 1, \] then \( 8e^2 + l^2 \) is equal to:
Let \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a>b \), be an ellipse whose eccentricity is \( \frac{1}{\sqrt{2}} \) and the length of the latus rectum is \( \sqrt{14} \). Then the square of the eccentricity of \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:
A parabola has vertex \((2, 3)\), equation of directrix is \(2x – y = 1\) and equation of ellipse is \(\frac {x^2}{a^2} +\frac {y^2}{b^2} =1\), \(e=\frac {1}{\sqrt 2}\) and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is
If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is.
If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:
The length of the latus rectum of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) \( (a>b) \) is \( \frac{8}{3} \). If the distance from the center of the ellipse to its focus is \( \sqrt{5} \), then \( \sqrt{a^2 + 6ab + b^2} = \text{?} \)
The eccentricity of the ellipse \[ p x^2 + 5y^2 = 80, \quad \text{where } p > 5, \] is \( \frac{\sqrt{3}}{2} \). Then the value of \( p \) is

The product of perpendiculars from the two foci of the ellipse $$ \frac{x^2}{9} + \frac{y^2}{25} = 1 $$ on the tangent at any point on the ellipse is: 

The end of a latus rectum of the ellipse \(3x^2 + 4y^2 = 12\) is lying in the third quadrant. If the normal drawn at \(L_1\) to this ellipse intersects the ellipse again at the point \(P(a,b)\), then find the value of \(a\):
The equations of the directrices of the ellipse \(9x^2 + 4y^2 - 18x - 16y - 11 = 0\) are:
S' is the focus of the ellipse \( \frac{x^2}{25} + \frac{y^2}{b^2} = 1, (b<5) \) lying on the negative X-axis and \( P(\theta) \) is a point on this ellipse. If the distance between the foci of this ellipse is 8 and \( S'P = 7 \), then \( \theta \) is:
If \( 4x - 3y - 5 = 0 \) is a normal to the ellipse \( 3x^2 + 8y^2 = k \), then the equation of the tangent at point (-2,m) is: