(1 point) Find an equation of the tangent line to the curve $x^{2 / 3}+y^{2 / 3}=10$ (an astroid) at the point $(27,1)$. \[ y= \] Preview My Answers Submit Answers

Step 1 ：Differentiate the given equation \(x^{2/3} + y^{2/3} = 10\) with respect to \(x\) to get \((2/3)x^{-1/3} + (2/3)y^{-1/3} * dy/dx = 0\).

Step 2 ：Solve this equation for \(dy/dx\) to find the slope of the tangent line at any point \((x, y)\), which gives \(dy/dx = -x^{-1/3} / y^{-1/3}\).

Step 3 ：Substitute \(x = 27\) and \(y = 1\) into the equation to get the slope of the tangent line at the point \((27, 1)\), which is \(dy/dx = -1/3\).

Step 4 ：Use the point-slope form of the equation of a line, \(y - y_0 = m(x - x_0)\), where \(m = -1/3\), \(x_0 = 27\), and \(y_0 = 1\), to get \(y - 1 = -1/3(x - 27)\).

Step 5 ：Solve this equation for \(y\) to get the equation of the tangent line, which is \(y = -1/3x + 10\).

Step 6 ：\(\boxed{y = -\frac{1}{3}x + 10}\) is the final answer.