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 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:
- 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:
- 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:
- The area of the region enclosed by the curve $ \left\{ (x, y) : 4x^2 + 25y^2 = 100 \right\} $ is:
Questions Asked in JEE Main exam
- The sum of all rational numbers in \( (2 + \sqrt{3})^8 \) is:
- JEE Main - 2025
- Binomial theorem
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
- Which of the following quantity has the same dimensions as \( \sqrt{\frac{\mu_0}{\epsilon_0}} \)?
- JEE Main - 2025
- Units, Dimensions and Measurements
- Number of solutions in the interval \( [-2\pi, 2\pi] \) for the equation: \[ 2\sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0 \]
- JEE Main - 2025
- Trigonometry
- Statement-I: Melting point of neopentane is greater than that of n-pentane.
Statement-II: Neopentane gives only one mono-substituted product.- JEE Main - 2025
- Isomerism
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}