Question:

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 x225+y2100=1\dfrac{x^2}{25}+\dfrac{y^2}{100}=1.

Updated On: Oct 21, 2023
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Solution and Explanation

The given equation is x225+y2100=1\dfrac{x^2}{25}+\dfrac{y^2}{100}=1 or  x252+y2102=1\dfrac{x^2}{5^2}+\dfrac{y^2}{10^2}=1

Here, the denominator of y2100\dfrac{y^2}{100} is greater than the denominator of x225\dfrac{x^2}{25}
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis. On comparing the given equation with x2b2+y2a2=1\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1, we obtain b=5b = 5 and a=10.a = 10.

 c=(a2 –b2)∴ c = √(a^2 – b^2)

=(10025)= √(100-25)

=75= √75

=53= 5√3
Therefore, 
The coordinates of the foci are (0,±53).(0, ±5√3) .
The coordinates of the vertices are (0,±10)(0, ±10)
Length of major axis = 2a=202a= 20
Length of minor axis = 2b=102b= 10

Eccentricity,e=ca=5310=32e = \dfrac{c}{a} = \dfrac{5√3}{10} = \dfrac{√3}{2}

Length of latus rectum = 2b2a=(2×52)10=(2×25)10=5.\dfrac{2b^2}{a} = \dfrac{(2×5^2)}{10} = \dfrac{(2×25)}{10} = 5.  

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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}