How do you find the coordinates of the center, foci, the length of the major and minor axis given #7x^2+3y^2-28x-12y=-19#?

Question

2345 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-29T05:38:27

Center is at #(2,2)#, Major axis length is # 2 sqrt 7 # , Minor axis length is # 2 sqrt3#, Focii are at # (2,4) and (2,0)# .

# 7 x^2+3 y ^2-28 x-12 y =-19# or

# 7 (x^2-4 x) +3( y ^2-4 y) =-19# or

# 7 (x^2-4 x +4) +3( y ^2-4 y+4) =28+12-19# or

# 7 (x-2)^2 +3( y-2)^2 =21# or

# (7(x-2)^2)/21+(3( y-2)^2)/21 =21/21# or

#(x-2)^2/3+( y-2)^2/7 =1 ; 7 >3# This is standard equation of

vertical ellipse with center at #(x_1=2,y_1=2) #and equation is

# (x-x_1)^2/b^2+(y-y_1)^2/a^2=1; b = sqrt 3, a= sqrt 7 #

Major axis length is #2 a= 2 sqrt 7 # , Minor axis length is

#2 b = 2 sqrt3 ; c^2= a^2-b^2= 7- 3 = 4 :. c= +- 2#

Focii are at #2,(2+2)and 2,(2-2) or (2,4) and (2,0)# on

major axis .

graph{7 x^2 +3 y^2 -28 x -12 y= -19 [-10, 10, -5, 5]} [Ans]