\(\frac{\sqrt3}{2}\)
\(\frac{1}{2\sqrt2}\)
\(\frac{1}{\sqrt2}\)
\(\frac{1}{2}\)
The correct answer is (C) : \(\frac{1}{\sqrt2}\)
Let \(P(2cosθ, √2sinθ )\) be any point on ellipse \(\frac{x²}{4} + \frac{y²}{2} = 1\)
and Q(4,3) and let (h, k) be the mid point of PQ then
\(h = \frac{2cosθ+4}{2}\)\(,k = \frac{\sqrt2Sinθ+3}{2}\)
\(∴ cosθ = h - 2 , sinθ = \frac{2k - 3}{\sqrt2}\)
\(∴ ( h - 2 )² + ( \frac{2k - 3}{\sqrt2} )^2 = 1\)
\(⇒ \frac{( x - 2 )²} {1} + \frac{( \frac{y - 3}{2} )^2 } {\frac{1}{2}} = 1\)
\(∴ e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt2}\)
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
Read More: Conic Section
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)]
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
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}