The area of the quadrilateral formed by the tangents at
the end points of latusrectum to the ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$, is
- 27/4 sq units
- 9 sq units
- 27/2 sq uni
- 27 sq units
The Correct Option is D
Approach Solution - 1
To find tangents at the end points of latusrectum , we
find ae
i.e. ae = $\sqrt {a^2-b^2}$ = $ \sqrt4 = 2$
and $\sqrt{b^2(1-e^2)}$ = $\sqrt{5\Big(1-\frac{4}{9}\Big)}$ = $\frac{5}{3}$
By symmetry, the quadrilateral is a rhombus
So, area is four times the area of the right angled
triangle formed by the tangent and axes in the 1st
quadrant.
$\therefore$ Equation of tangent at $\Big(2,\frac{5}{3}\Big)$ is
$\frac{2}{9} x + \frac{5}{3}.\frac{y}{5}$ = 1 $\Rightarrow \frac{3}{\frac{9}{2}}+\frac{y}{3}$ = 1
$therefore$ Area of quadrilateral ABCD
= 4 [area of $\Delta$ AOB]
= 27 sq units
Approach Solution -2
Ans. A quadrilateral, which has four sides, four angles, and four vertices, is a two-dimensional form. A quadrilateral's sides might have equal or unequal lengths based on its characteristics and form. A quadrilateral can have any number of distinct sides, but the total of all these sides is always equal to 360°. The most common shapes for quadrilaterals are parallelograms, squares, rectangles, rhombuses, and trapezoids (trapeziums). A quadrilateral is a figure with four straight lines; the total of all the angles of a quadrilateral is 360°. A quadrilateral can be a parallelogram, square, rectangle, rhombus, trapezium, or kite.
Ellipse is a plane curve that surrounds two focal points, in such a way that for all points on the curve, the sum of two distances to the focal point is a constant. It resembles a circle, which is a special type of ellipse in which both the focal points are the same.
- An ellipse is a closed plane curve, obtained by a point that moves in such a way that the sum of their distances from any two fixed points remains a constant.
- A plane section from a right circular cone is a closed curve.
- The fixed points are called ‘foci’ which are surrounded by the curve.
The equation of the ellipse is given by: x2/a2 + y2/b2 = 1
- The major axis and minor axis define the area of an oval-shaped ellipse.
- The eccentricity that demonstrates the elongation of the ellipse is denoted by the variable “e”.
- Ellipse is part of the conic section. For example, the Parabola and the Hyperbola
- Parabola is open in shape and unbounded. Generally, an ellipse is defined by its equation.
Top Questions on Ellipse
- 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:
- 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 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:
- 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 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:
Questions Asked in JEE Advanced exam
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Waves and Oscillations
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- If $$ \alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx, $$ then the value of $ \sqrt{7} \tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) $ is.
(Here, the inverse trigonometric function $ \tan^{-1} x $ assumes values in $ \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $.)- JEE Advanced - 2025
- Integral Calculus
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Concepts Used:
Ellipse
Ellipse Shape
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Properties
- Ellipse has two focal points, also called foci.
- The fixed distance is called a directrix.
- The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
- The total sum of each distance from the locus of an ellipse to the two focal points is constant
- Ellipse has one major axis and one minor axis and a center
Read More: Conic Section
Eccentricity of the Ellipse
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
Area of an ellipse
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Position of point related to Ellipse
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}