Find the equation for the ellipse that satisfies the given conditions: Vertices \((±5,0),\)foci (±4,0)
Find the equation for the ellipse that satisfies the given conditions: Vertices \((±5,0),\)foci (±4,0)
Solution and Explanation
Vertices\( ( ±5, 0)\), 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, \(a = 5\) and c = 4.
It is known that,
\(a2 = b2 + c2.\)
So, \(5^2 = b^2 + 4^2\)
\(25 = b^2 + 16\)
\(b^2 = 25 – 16\)
\(b = √9\)
\(= 3\)
∴ The equation of the ellipse is \(\dfrac{x^2}{5^2} + \dfrac{y^2}{3^2} = 1 \)
\(\dfrac{ x^2}{25} + \dfrac{y^2}{9} = 1\) (Ans)
Top Questions on Ellipse
- Let the ellipse \( E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a > b \) and \( E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \), \( A < B \), have the same eccentricity \( \sqrt{\frac{1}{3}} \). Let the product of their lengths of latus rectums be \( \sqrt{\frac{32}{3}} \) and the distance between the foci of \( E_1 \) be 4. If \( E_1 \) and \( E_2 \) meet at \( A \), \( B \), \( C \), and \( D \), then the area of the quadrilateral ABCD equals:
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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}