Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse \(\dfrac{x^2}{49}+\dfrac{y^2}{36}=1\)
Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse \(\dfrac{x^2}{49}+\dfrac{y^2}{36}=1\)
Solution and Explanation
The given equation is \(\dfrac{x^2}{49}+\dfrac{y^2}{36}=1\) = 1 or \(\dfrac{x^2}{7^2}+\dfrac{y^2}{6^2}=1\)
Here, the denominator of \(\dfrac{x^2}{49}\) is greater than the denominator of \(\dfrac{y^2}{36}. \)
Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.
On comparing the given equation with \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\) , we obtain \(a = 7\) and \(b = 6.\)
∴ \(c = √(a^2 – b^2)\)
\(= √(49-36)\)
\(= √13\)
Therefore,
The coordinates of the foci are \(( ±√13, 0). \)
The coordinates of the vertices are \(( ±7, 0). \)
Length of major axis = \(2a= 14\)
Length of minor axis = \(2b= 12\)
Eccentricity,\( e = c/a = √13/7\)
Length of latus rectum = \(\dfrac{2b^2}{a} = \dfrac{(2×6^2)}{7} = \dfrac{(2×36)}{7} = \dfrac{72}{7}\)
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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}