Find the equation of tangent and normal of curve x^2/3+y^2/3=a^2/3 at (h,k)

Question

Arun · Answer

Dear student To find the tangents to a give curve,differentiate the the give curve.... On differentiating...we get slope of the tangent.. Differentiating with respect to x, 2/3(x-1/3) + 2/3(y-1/3) dy/dx = 0 dy/dx = -x-1/3/y-1/3 NOW FIND slope at (h,k) and then write equation of tangent

Vikas TU · Answer

Dear student Please follow this link https://www.askiitians.com/iit-jee-differential-calculus/tangents-and-normal/ Good Luck

Aditya Gupta · Answer

the correct answer is replace x^2 by xh, and y^2 by yk so, eqn of tangent is xh/3+yk/3=a^2/3 (slope = – h/k) normal can be written as y – k= m(x – h) where m*( – h/k)= – 1 or m= k/h or yh – kh= kx – kh or y= kx/h kindly approve :))

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Find the equation of tangent and normal of curve x^2/3+y^2/3=a^2/3 at (h,k)

Find the equation of tangent and normal of curve x^2/3+y^2/3=a^2/3 at (h,k)

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