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Ex 9.3, 18 At any point (𝑥 , 𝑦) of a curve , the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (−4 , −3) . Find the equation of the curve given that its passes through(−2 , 1) Slope of tangent to the curve = 𝒅𝒚/𝒅𝒙 Slope of line segment joining (x, y) & (−4, −3) = (𝑦2 − 𝑦1)/(𝑥2 −𝑥1) = (−𝟑 − 𝒚)/(−𝟒 − 𝒙) = (−(𝑦 + 3))/(−(𝑥 + 4)) = (𝒚 + 𝟑)/(𝒙 + 𝟒) Given, at point (x, y). Slope of tangent is twice of line segment 𝒅𝒚/𝒅𝒙 = 2((𝒚 + 𝟑)/(𝒙 + 𝟒)) 𝑑𝑦/(𝑦 + 3) = (2 𝑑𝑥)/(𝑥 + 4) Integrating both sides ∫1▒〖𝑑𝑦/(𝑦 + 3) " " 〗= 2∫1▒〖" " ( 𝑑𝑥)/(𝑥 + 4)〗 log (y + 3) = 2 log (x + 4) + log C log (y + 3) = log (x + 4)2 + log C log (y + 3) − log (x + 4)2 = log C log (𝑦 + 3)/(𝑥 + 4)^2 = log C (𝒚 + 𝟑)/(𝒙 + 𝟒)^𝟐 = C The curve passes through (−2, 1) Put x = −2 & y = 1 in (1) (1 + 3)/(−2 + 4)^2 = C C = 4/(2)^2 = 4/4 C = 1 Put value c = 1 in equation (1) (𝑦 + 3)/(𝑥 + 4)^2 = 1 y + 3 = (x + 4)2 Hence the equation of the curve is y + 3 = (x + 4)2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo