$P$ is a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ with foci $F_1$ and $F_2$ . If $A$ is the area of the triangle $PF_1F_2$. then the maximum value of $A$ is
- $\frac{e}{ab}$
- $\frac{ab}{e}$
- $aeb$
- $\frac{ab}{a}$
The Correct Option is C
Solution and Explanation
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:
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Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
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Identify the anomers from the following.
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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}