Question

Write the standard equation of the ellipse:

1. F1 (-4,0) and F2 (4,0), passing through (4,1)

Answer 1

the centre of this one is (0,0) and the major axis lies horizontally, so
x^2/a^2 + y^2/b^2 = 1 **

for such an ellipse: a^2 = b^2 + c^2, but c = 4
a^2 = b^2 + 16

sub back into **
x^2/(b^2+16) + y^2/b^2 = 1
but (4,1) lies on it, so
16/(b^2+16) + 1/b^2 = 1
16b^2 + b^2+16 = b^2(b^2 + 16)
17b^2 + 16 = b^4 + 16b^2

b^2 = (1 ± √65)/2 = (1 + √65)/2 since b^2 > 0
then a^2 = (33 + √65)/2

x^2 / ( (33 + √65)/2) + y^2/( (1 + √65)/2) = 1

Answer 2

thank you

Answer 3

16b^2 + b^2+16 = b^2(b^2 + 16)
how did this become this 17b^2 + 16 = b^4 + 16b^2 where did 17 come from?

Answer 4

because he/she add the 16b^2 and b^2
b^2 is 1b^2 and they have the same variable and same exponent so you should combine them that why it become 17b^2 :)

Answer 5

Hi, I'm confused. May I ask where she got:
[ b^2(b^2+16) ] after this part:
[ 16/(b^2+16) + 1/b^2 = 1 ] ?

Answer 6

The b^2 and b^2+16 were multiplied on the right side of the equation, making it b^2(b^2 + 16)

Answer 7

where did √65 come from?
after 17b^2 + 16 = b^4 + 16b^2
what did he /she do next?