Question

Find the equation of the ellipse having it's centre at point (2,-3) one focus at (3,-3) and one vertex at (4,-3) URGENT

Answer 1

Answer:

(x - 2)²/4 + (y + 3)² = 1

Step-by-step explanation:

Hi,

Given center of the ellipse , O (2, -3)

Let Focus of ellipse, S be (3, -3)

Let One Vertex of ellipse, A be ( 4, -3).

Now, we know the line joining the center to the focus is the equation of major axis,

hence y = -3 is equation of major axis.-------(1)

Minor axis is the line perpendicular to Major axis and passing through the center,

hence equation of line perpendicular to major axis (y = -3) will be of form x= k but it passes through center, hence  = 2.

Hence x = 2 is the equation of minor axis.------(2)

Distance between the center and the vertex is the semi major axis,a whch is 2, hence a = 2----(3)

Distance between the center and the focus of the ellipse is ae which is 1

=> ae = 1

from (3), we get 2e = 1

=> e = 1/2

We know that b² = a²(1 - e²)

=> b² = 4(3/4)

=> b = √3.

NOTE:Equation of ellipse when major axis equation, minor axis equation and their corresponding lengths are known,

(⊥ distance to minor axis)²/a² + (⊥ distance to major axis)²/b² = 1

=> (x - 2)²/4 + (y + 3)² = 1 is the required equation of an ellipse.

Hope, it helped !

Answer 2

Answer:

Step-by-step explanation:

Concept:

In an ellipse,

Distance between centre and focus is

CF = ae

Distance between centre and vertex is

CA = a

Distance between centre and focus,

CF=$\sqrt{(2-3)^2+(-3+3)^2}$

CF= 1

That is, ae = 1.....(1)

Distance between centre and vertex,

CA=$\sqrt{(2-4)^2+(-3+3)^2}$

CA= 2

That is, a = 2........(2)

using (2) in (1), we get

e=1/2

$b^2=a^2(1-e^2)\\\\b^2=4(1-\frac{1}{4})\\\\b^2=4(\frac{3}{4})\\\\b^2=3$

since the major axis is along x axis,

the equation of the ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\\\\\frac{(x-2)^2}{4}+\frac{(y+3)^2}{3}=1$