$B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S'$. If $?SBS'$ is a right angle, then the eccentricity of the ellipse is
- $\frac{1}{2}$
- $\frac{1}{\sqrt{2}}$
- $\frac{2}{3}$
- $\frac{1}{3}$
The Correct Option is B
Solution and Explanation
Slope of $SB, m_1 = \frac{b - 0}{0 - ae} =-\frac{b}{ae}$
and slope of $S^{\prime} B, m_{2}=\frac{b-0}{0-(-a e)}=\frac{b}{a e}$
Since, $\angle S B S^{\prime}$ is a right angle.
$\therefore m_{1} m_{2} =-1 $
$\Rightarrow \frac{-b}{a e} \times \frac{b}{a e} =-1 $
$\Rightarrow b^{2} =a^{2} e^{2} $
$\Rightarrow \frac{b^{2}}{a^{2}} =e^{2} $
$\Rightarrow 1-e^{2} =e^{2} $
$\Rightarrow 2 e^{2} =1$
$\Rightarrow e^{2}=\frac{1}{2} $
$\Rightarrow e=\pm \frac{1}{\sqrt{2}} $
$ \Rightarrow e=\frac{1}{\sqrt{2}} $
or $ e=-\frac{1}{\sqrt{2}}$
Top Questions on Ellipse
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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:
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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}