asymptotes of the curve x^2y^2=a^2(x^2+y^2) form a square of side 2a
Answers
Answered by
8
axx2+bxy2=1axx2+bxy2=1
Rearranging this into a nicer form:
y2=−b2∗x2a2−x2y2=−b2∗x2a2−x2
A vertical asymptote will occur when the denominator = 0.Therefore :
x=+/−ax=+/−a
For the Horizontal asymptope take the limit as x → infinity
y2=limx→infinity−b2∗x2x2a2x2−1y2=limx→infinity−b2∗x2x2a2x2−1
y2=limx→infinityb21y2=limx→infinityb21
Therefore Horizontal assomtyopes at :
y=+/−b
Rearranging this into a nicer form:
y2=−b2∗x2a2−x2y2=−b2∗x2a2−x2
A vertical asymptote will occur when the denominator = 0.Therefore :
x=+/−ax=+/−a
For the Horizontal asymptope take the limit as x → infinity
y2=limx→infinity−b2∗x2x2a2x2−1y2=limx→infinity−b2∗x2x2a2x2−1
y2=limx→infinityb21y2=limx→infinityb21
Therefore Horizontal assomtyopes at :
y=+/−b
Similar questions