\(\frac{\sqrt3}{2}\)
\(\frac{1}{2\sqrt2}\)
\(\frac{1}{\sqrt2}\)
\(\frac{1}{2}\)
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
\(h = \frac{2cosθ+4}{2}\)\(,k = \frac{\sqrt2Sinθ+3}{2}\)
\(∴ cosθ = h - 2 , sinθ = \frac{2k - 3}{\sqrt2}\)
\(∴ ( h - 2 )² + ( \frac{2k - 3}{\sqrt2} )^2 = 1\)
\(⇒ \frac{( x - 2 )²} {1} + \frac{( \frac{y - 3}{2} )^2 } {\frac{1}{2}} = 1\)
\(∴ e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt2}\)
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
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}