Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 9.3

Ex 9.3, 1
Important

Ex 9.3, 2

Ex 9.3, 3

Ex 9.3, 4 Important

Ex 9.3, 5

Ex 9.3, 6

Ex 9.3, 7 Important

Ex 9.3, 8

Ex 9.3, 9 Important

Ex 9.3, 10 Important

Ex 9.3, 11 Important

Ex 9.3, 12

Ex 9.3, 13

Ex 9.3, 14

Ex 9.3, 15 Important

Ex 9.3, 16

Ex 9.3, 17 Important You are here

Ex 9.3, 18

Ex 9.3, 19 Important

Ex 9.3, 20 Important

Ex 9.3, 21

Ex 9.3, 22 Important

Ex 9.3, 23 (MCQ)

Last updated at May 29, 2023 by Teachoo

Ex 9.3, 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