Find the equation for the ellipse that satisfies the given conditions:Vertices (±6,0),foci (±4,0)
Solution and Explanation
Vertices \(( ±6, 0), \) foci \(( ±4, 0) \)
Here, the vertices are on the \(x-axis. \)
Therefore, the equation of the ellipse will be of the form \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\), where ‘a’ is the semi-major axis.
Accordingly, \(a = 6\)\(\) , \(c= 4. \)
It is known that
\(a^2 = b^2 + c^2.\)
\(6^2 = b^2+4^2\)
\(36 = b^2 + 16\)
\(b^2 = 36 – 16\)
\(b = √20\)\(\)
∴ The equation of the ellipse is \(\dfrac{x^2}{6^2} + \dfrac{y^2}{(√20)^2} = 1\) or \(\dfrac{x^2}{36} + \dfrac{y^2}{20} = 1\)
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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}