Question

Find all the asymptotes of x' + x²y – xy2 - y2 + 2xy + 2y2 – 3x + y = 0.
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Answer 1

Answer:

Step-by-step explanation:

Rewrite the given equation as x2y – xy2 – xy + y2 + x – y = 0 as yx2 + (1 – x)y2 – xy + x – y = 0 Clearly asymptotes parallel to X-axis and Y-axis are y = 0 and x = 1 respectively. For oblique asymptotes: φ3(m) = m – m2 and φ2(m) = –m + m2 implying φ'3(m) = 1 – 2m so that c = Therefore, equation of asymptotes y = mx + c becomes y = 0 , y = x. Now the joint equation of the asymptotes, y(x – 1)(y – x) = 0 or x2y – xy2 – xy + y2 = 0 Clearly given equation of the curve, i.e. (x2y – xy2 – xy + y2 ) + (x – y) = 0 is expressible like Fn + Fn – 2 = 0 where in we have obtained, Fn = 0, i.e. x2y – xy2 – xy + y2 = 0 as the joint equation of the asymptotes. and c = - φ2(m)/φ3(m) = - ( - 2am2)/ 2m = am = ± a Therefore the two oblique asymptotes are y = x + a and y = –x – a.

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