Question

निदर्शी उदाहरण (Illustrative Exar 11. निम्नलिखित वक्र की अनन्तस्पर्शियाँ ज्ञात कीजिए : 13-2xy + xy2 +x2 –xy + 2 = 0. चक्र का सपी गिळित है [F =​

Answer 1

Here the line y = 0 is the asymptote parallel to X-axis whereas there is no asymptote parallel to Y-axis. For Oblique Asymptotes: In the given equation of curve, expression containing the third degree terms is y3 + x2y + 2xy2 Thus, φ3(m) = m3 + 2m2 + m (by taking y = m, x = 1) so that φ'3(m) = 3m2 + 4m + 1 and φ"3(m) = 6m + 4 Likewise, φ2(m) = 0, φ1(m) = –m φ3 = 0 ⇒ m3 + 2m2 + m = 0 or m = –1, –1, 0 Now for equal values of m in φn(m), corresponding values of ‘c’ are obtained from ⇒ c2/2(6m + 4) + c.o - m = 0 or c2 = m/ 3m + 2 for m = -1, c2 = m/ 3m + 2 = -1/ -3 + 2 = 1 implying c = ± 1 and for m = 0, already we had obtained the parallel asymptote. Therefore, the asymptotes are y = 0, y = –x + 1, y = –x – 1.Read more on Sarthaks.com - https://www.sarthaks.com/495998/find-the-asymptotes-of-the-curve-y-3-x-2y-2xy-2-y-1-0

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