# How do you find the equation of the ellipse that passes through points (6,4) and (-8,3) ?

##### 1 Answer

#### Answer:

#x^2/100+y^2/25=1#

#### Explanation:

Two Points are given. The center is not given.

We shall take

The equation of the ellipse is -

#(x-h)^2/a^2+(y-k)^2/b^2=1#

Plug in the values of center

#(x-0)^2/a^2+(y-0)^2/b^2=1#

This is the equation of the ellipse having center as

#x^2/a^2+y^2/b^2=1#

The given ellipse passes through points

First plugin the values

#6^2/a^2+4^2/b^2=1#

#36/a^2+16/b^2=1# ------------(1)

Next Plugin the values

#(-8)^2/a^2+3^2/b^2=1#

#64/a^2+9/b^2=1# -----------------(2)

#36/a^2+16/b^2=1# ------------(1)

#64/a^2+9/b^2=1# -----------------(2)

We shall rewrite the equations as -

#36 1/a^2+16 1/b^2=1# ------------(1)

#64 1/a^2+9 1/b^2=1# -----------------(2)

Let

Then these two equations become

#36m+16n =1# ------------(1)

#64m+9n=1# -----------------(2)

Now we can easily solve

#36m+16n =1# ------------(1)#xx9#

#64m+9n=1# -----------------(2)#xx16#

#324m+144n =9# ------------(1)

#1024m+144n=16# -----------------(2)

Subtract (2) from (1)

#-700m=-7#

#m=(-7)/(-700)=1/100#

Plugin this in any one of the euations

#36 * 1/100+16n =1#

#16n=1-36/100=(100-36)/100=16/25#

#n=16/25*1/16=1/25#

#m=1/a^2=1/100#

#n=1/b^2=1/25#

#x^2/a^2+y^2/b^2=1#

This equation can also be written as -

#1/a^2*x^2+1/b^2*y^2=1#

Now you can easily plugin the values of

#1/100*x^2+1/25*y^2=1#

#x^2/100+y^2/25=1#