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 $ 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?
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- 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:
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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}