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Find the equations of tangent and normal to the curve xy = 10Vh T |
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Answer» Equation of tangent is given by = (y-y1)= M(x-x1)M=dy/dx of the curvexy=10y=10/xdy/dx = (-10/x^2 ) at x = 2M= -10/4=-5/2 so we get the slope nowy1=5 x1 =2y-y1=m(x-x1) be equation of tangenty-5=-5/2(x-2)y+5/2x -10 = 02y + 5x - 20 =0 is the equation of tangent now for equation of normalwe need Mn by using this equationM*Mn = -1Mn *-5/2 =-1Mn =2/5 for equation of tangenty-y1 =Mn (x-x1)y-5 = 2/5(x-2)y-2/5 x -5 + 4/5 = 05y - 2x -21 = 0 be the equation of tangent |
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