# Ex 9.4, 17 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.4, 17 Find the equation of a curve passing through the point 0 , −2 , given that at any point 𝑥 , 𝑦 on the curve , the product of the slope of its tangent and 𝑦 coordinate of the point is equal to the 𝑥 coordinate of the point . Slope of tangent to the curve = 𝑑𝑦𝑑𝑥 Given at any point (x, y) y 𝑑𝑦𝑑𝑥 = x y dy = x dx Integrating both sides 𝑦 𝑑𝑦= 𝑥 𝑑𝑥 𝑦22 = 𝑥22 + C The curve passes through point (0, −2) Put x = 0 & y = −2 −222 = 022 + C 42 = C C = 2 Put value in equation 𝑦22 = 𝑥22 + 2 𝑦22 = 𝑥2 + 42 𝑦2 = 𝑥2 + 4 𝑦2 − 𝑥2= 4 Hence, equation of the curve is 𝒚𝟐 − 𝒙𝟐= 4

Chapter 9 Class 12 Differential Equations

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.