Solution 2





In this page 'Solution 2' we see step by step methods to find the foci, centre, eccentricity and latus rectum from the given equation of the ellipse.

Before solving the practice problems let us recall the important points to find the foci, latus rectum and major axis.

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.

Example :

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.

The formula to find the foci is (ae,0) and (-ae,0) if the major axis is horizontal axis. (0, be) and (0,-be) if the major axis is the vertical axis when the centre is the origin.

The formula to find the latus rectum is 2a²/b.

To find the eccentricity from 'a' and 'b' is

                                   a²=b²(1-e²)    Solution 2

Practice problem 1: 

Find the eccentricity, foci and latus rectum of the ellipse 9x²+4y²=36.

Solution for practice problem 1: 

        The given equation of ellipse is 9x²+4y²=36.

        Dividing the equation by 36, we get

                        x²/4   +   y²/9     =   1.

       Here a =2 and b= 3, i.e., a<b

       The major axis is Y axis (vertical axis)

                              a² = b²(1-e²)

                              4  =  9(1-e²)

                              4  =   9-9e²

                              e² =    5/9

                               e  =   (√5)/3

                      Foci are (0,be) and (0,-be)

                                  (0,3[(√5)/3]) and (0,-3[(√5)/3])

                                   (0,√5) and (0,-√5)

                       Latus rectum =2a²/b  =  2 (4/3) = 8/3

The eccentricity is (√5)/3,  foci are  (0,√5) and (0,-√5) and the latus rectum is 8/3    


Practice problem 2:

            Find the foci, centre, eccentricity and latus rectum of the ellipse whose equation is 25x²+9y²-150x-90y+225=0

Solution for practice problem 2:

The given equation of the ellipse is 25x²+9y²-150x-90y+225=0

 (25x²-150x)+(9y²-90y)+225 = 0

  25(x²-6x+9) + 9(y²-10y+25)  =  225

   25(x-3)² + 9(y-5)²   =  225

Dividing by 225

   (x-3)²/9   +   (y-5)²/25       =   1

    So centre is (3,5)

     a=3 and b= 5 so a<b.

    The major axis is Y(vertical) axis.

 The eccentricity is e² = 1-(a²/b²)

                               = 1-(9/25)

                               =   16/25

                          e   =   4/5

Foci are (0,be) and (0,-be) when the centre is the origin.

Foci when the centre is the origin (0,4) and (0,-4)

Foci with respect to the new centre (3,5) are

               (3,9) and (3,1)

Latus rectum is 2a²/b = 2.(9/5) = 18/5.

The centre is (3,5), the foci are (3,9) and (3,1), eccentricity is 4/5 and the latus rectum is 18/5.                    

Problems for practice:      

  1. Find the foci, eccentricity and latus rectum for the equation of the ellipse 3x²+4y²-12x-8y+4=0.
  2. Find the foci, centre, eccentricity and latus rectum for the equation of the ellipse 9x²+25y²-18x-100y-116=0.

Parents and teachers can guide the students to follow the solutions in this page 'Solution 2', step by step. Students can practice the problems in the methods discussed above and try to solve the problems given below on their own. If you have any doubt you can contact us through mail, we will help you to clear your doubts.         





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