Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

Find the equation of the ellipse that has its foci at (2,1) and (2,-7) and b= 2

Sagot :

Foci: (2,1) and (2,-7)

[tex]F_1=\text{ (0+ x', 0 + y')}[/tex]

The center of the ellipse is 2, 3

[tex]\begin{gathered} F_1=\text{ (0+ 2, 1 + 3)}=>F_1(0,4) \\ F_2=(0+2,-7+3)\Rightarrow F_2(0,-4) \\ \text{Hence, c=4} \end{gathered}[/tex]

[tex]F_2=\text{ (0+x', 0 + y')}[/tex]

From

[tex]\begin{gathered} c^2=a^2-b^2 \\ a^2=c^2+b^2 \\ c=4,\text{ b=2} \\ a^2=4^2+2^2=16+4=20 \\ a^2=20 \end{gathered}[/tex]

The general equation of an ellipse with center ( x' ,y')

where, x'=2 and y' =3

[tex]\frac{(x-x^{\prime})^2}{b^2}+\frac{(y-y^{\prime})^2}{a^2}=1[/tex][tex]\begin{gathered} \frac{(x-2)^2}{4}+\frac{(y-3)^2}{20}=1 \\ \frac{5(x-2)^2+(y-3)^2}{20}=1 \end{gathered}[/tex][tex]\begin{gathered} 5(x-2)^2+(y-3)^2=20 \\ 5(x^2-4x+4)+(y^2-6y+9)=20 \\ 5x^2-20x+20+y^2-6y+9=20 \\ 5x^2+y^2-20x-6y+29-20=0 \\ 5x^2+y^2-20x-6y+9=0 \end{gathered}[/tex]

Hence the equation of the ellipse is

[tex]5x^2+y^2-20x-6y+9=0[/tex]