In this page 'Equation of ellipse' we are going to examples which describes how to get the equation of the ellipse from the given foci, eccentricity and directrix.
Find the equation of the ellipse whose focus is (-1,1), eccentricity is 1/2 and whose directrix is x-y+3=0.
The focus is (-1,1).
Directrix is x-y+3=0 and e =1/2.
Let P(x₁,y₁) be any point on the ellipse, then SP² = e²PM² where PM is perpendicular distance from x-y+3=0
(x₁+1)² + (y₁-1)² = 1/4 [(x₁-y₁+3)/√(1+1)]²
8(x₁+1)² +8 (y₁-1)² = (x₁-y₁+3)²
Locus of (x₁,y₁) i.e., the equation of the required ellipse is
Find the equation of the ellipse whose foci are (2,0) and (-2,0) and eccentricity is 1/2.
S(ae,0) and S'(ae,0) are the foci of the ellipse (x²/a²)+(y²/b²) =1.
The foci are (2,0) and (-2,0) and e=1/2
ae = 2 and e = 1/2
a= 4 and a² =16
The center C is the mid point of SS' is (0,0). S and S' are on the x axis. So the equation of the ellipses of the form (x²/a²)+(y²/b²) =1
b² = a²(1-e²) = 16(1-1/4) = 12.
The equation of the required ellipse is (x²/16)+(y²/12) =1.
We know that the equation of the ellipse is (x²/a²)+(y²/b²) =1, where a is the major axis (which is horizontal X axis), b is the minor axis and a>b here.
If the equation is ,(x²/b²)+(y²/a²) =1 then here a is the major axis which is vertical Y axis, b is the minor axis and a>b.
In the equation (x²/16)+(y²/9) =1 we have a =4 and b=3. a>b, so the major axis is X axis.
In the equation (x²/16)+(y²/25) =1, we have a =4 and b=5 and >a.
So the major axis is the vertical axis that is Y axis
1. Find the equation of the ellipse whose focus is (1,2), directrix is 2x-3y+6=0 and the eccentricity is 2/3. Solution
2. Find the equation of the ellipse whose foci are (4,0) and (-4,0) and e =1/3. Solution
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