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Ex 9.4, 17 - Find equation: (0, -2), product of slope - Ex 9.4

Ex 9.4, 17 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 17 - Chapter 9 Class 12 Differential Equations - Part 3


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 = 𝑑𝑦/𝑑𝑥 Given at any point (x, y), product of slope of its tangent and y-coordinate is equal to x-coordinate of the point Therefore, y 𝑑𝑦/𝑑𝑥 = x y dy = x dx …(1) Integrating both sides ∫1▒〖𝑦 𝑑𝑦=∫1▒〖𝑥 𝑑𝑥 〗 〗 𝑦^2/2 = 𝑥^2/2 + C The curve passes through point (0, −2) Putting x = 0 & y = −2 in equation (−2)^2/2 = 0^2/2 + C 4/2 = C C = 2 Putting back value of C in equation 𝑦^2/2 = 𝑥^2/2 + 2 𝑦^2/2 = (𝑥^2 + 4)/2 𝑦^2 = 𝑥^2 + 4 𝑦^2 − 𝑥^2= 4 Hence, equation of the curve is 𝒚^𝟐 − 𝒙^𝟐= 4

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.