The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{81}=1$ and the co-ordinate axes is :
- 12
- 18
- 26
- 36
The Correct Option is D
Solution and Explanation
equation of tangent $\frac{x}{4} \cos \theta+\frac{y}{9} \sin \theta=1$
Let $ A \& B$ are point of intersection of tangent at $P$ with co-ordinate axes.
$A \left(\frac{4}{\cos \theta}, 0\right) B \left(0, \frac{9}{\sin \theta}\right)$
Area of $\Delta OAB =\frac{1}{2}\left(\frac{4}{\cos \theta}\right)\left(\frac{9}{\sin \theta}\right)=\frac{36}{\sin 2 \theta}$
$(\text { Area })_{\min }=36$ as $\sin 2 \theta=1$
Top Questions on Ellipse
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 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 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:
Questions Asked in JEE Main exam
- The system of linear equations
$x + y + z = 6$
$2x + 5y + az = 36$
$x + 2y + 3z = b$
has- JEE Main - 2026
- Matrices and Determinants
- The displacement of a particle executing simple harmonic motion with time period \(T\) is expressed as \[ x(t)=A\sin\omega t, \] where \(A\) is the amplitude of oscillation. If the maximum value of the potential energy of the oscillator is found at \[ t=\frac{T}{2\beta}, \] then the value of \(\beta\) is ________.
- JEE Main - 2026
- Waves and Oscillations
- A complex number 'z' satisfy both \(|z-6|=5\) & \(|z+2-6i|=5\) simultaneously. Find the value of \(z^3 + 3z^2 - 15z + 141\).
- JEE Main - 2026
- Algebra
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$)
- JEE Main - 2026
- Rotational Mechanics
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
- JEE Main - 2026
- Rotational motion
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}