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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)
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-1/3NOW FIND slope at (h,k) and then write equation of tangent
Dear student Please follow this link https://www.askiitians.com/iit-jee-differential-calculus/tangents-and-normal/Good Luck
the correct answer is replace x^2 by xh, and y^2 by ykso, eqn of tangent is xh/3+yk/3=a^2/3 (slope = – h/k)normal can be written asy – k= m(x – h)where m*( – h/k)= – 1or m= k/hor yh – kh= kx – khor y= kx/hkindly approve :))
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