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
Write the standard equation of the ellipse:
1. F1 (-4,0) and F2 (4,0), passing through (4,1)
7 answers
thank you
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?
how did this become this 17b^2 + 16 = b^4 + 16b^2 where did 17 come from?
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 :)
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 :)
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 ] ?
[ b^2(b^2+16) ] after this part:
[ 16/(b^2+16) + 1/b^2 = 1 ] ?
The b^2 and b^2+16 were multiplied on the right side of the equation, making it b^2(b^2 + 16)
where did √65 come from?
after 17b^2 + 16 = b^4 + 16b^2
what did he /she do next?
after 17b^2 + 16 = b^4 + 16b^2
what did he /she do next?
1 answer