Question

How do we find the eccentric angle of ellipse \[\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{4} = 1?\]

Answer 1

Hint: An ellipse is the locus of all those points in a plane such that the sum of their distances from two fixed points in the plane is constant. The fixed point is known as foci and the fixed line is directrix and the constant ratio is the eccentricity of the ellipse. Eccentricity is a factor of the ellipse which demonstrates the elongation of it and is denoted by “e”.
The equation of ellipse is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]
The eccentricity of an ellipse is the ratio of the distance between the centre of the ellipse and each focus to the length of the semi major axis.
The eccentric angle of a point on an ellipse with semi major axis of length \[a\] and semi minor axis of length \[b\] is the angle \[\theta \] , which is in the form for a point \[\left( {x,y} \right)\] ,
 \[
  x = a\cos \theta \\
  y = b\sin \theta \;
 \]
Therefore \[\theta \] \[ = \] \[{\tan ^{ - 1}}\left( {\dfrac{{ay}}{{bx}}} \right)\]
Complete step-by-step answer:
Given the equation of ellipse is \[\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{4} = 1\]
We know the general equation of ellipse is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] ,
Where \[a\] is the length of the semi major axis and \[b\] is the length of the semi minor axis.
Comparing the given equation with the general equation of an ellipse we get
 \[{a^2} = 16\] and \[{b^2} = 4\]
Therefore \[a = 4,b = 2\] . Since the square root of \[16\] is \[4\] and the square root of \[4\] is \[2\] .
We know for a point \[\left( {x,y} \right)\] ,
 \[
  x = a\cos \theta \\
  y = b\sin \theta \;
 \]
 \[\therefore \] \[
  x = 4\cos \theta \\
  y = 2\sin \theta \;
 \]
Where \[\theta \] is the eccentric angle .
Dividing \[y\] by \[x\] , we get
 \[\dfrac{y}{x} = \dfrac{{2\sin \theta }}{{4\cos \theta }}\]
 \[
   \Rightarrow \dfrac{y}{x} = \dfrac{1}{2}\tan \theta \\
   \Rightarrow \tan \theta = \dfrac{{2y}}{x} \\
 \] ;since \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \]
 \[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2y}}{x}} \right)\]
Hence the eccentric angle = \[{\tan ^{ - 1}}\left( {\dfrac{{2y}}{x}} \right)\]
So, the correct answer is “ \[{\tan ^{ - 1}}\left( {\dfrac{{2y}}{x}} \right)\] ”.

Note: Conic section is a curve obtained as the intersection of the surface of the cone with a plane . In other words any curve produced by the intersection of the plane and right circular cone .The three types of conic sections are :- Parabola , Ellipse , Hyperbola. To find the eccentric angle the coordinate should be known to us. We should not forget to take out the square root of the number which we get by comparing the denominator of the general equation and the given equation. Should have a clear concept about ellipse and its semi minor and semi major axis and also about its eccentricity.