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Find the equations of the tangent and normal to the curve `x^(2/3)+y^(2/3)=2`at (1, 1) |
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Answer» Equation of the given curve, `x^(2/3)+y^(2/3) = 2` Differentiating it w.r.t. x, `2/3x^(-1/3)+2/3y^(-1/3)dy/dx = 0` `=>dy/dx = -(x/y)^(-1/3) = -(y/x)^(1/3)` `:. ` Slope of the tangent at point `(1,1) = m_1 = dy/dx = -(1)^(1/3) = -1` Let slope of the normal is `m_2`. then, `m_1*m_2 = -1 => -1*m_2 = -1 => m_2 = 1` Now, equation of the tangent, `y-1 = -1(x-1) => x+y-2 = 0` Equation of the normal, `y-1 = x-1 => x-y = 0.` |
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