Question 3 of 5 The given point is on the curve. Find the lines that are (a) tangent and (b) normal to the curve at the given point. \[ 2 x^{2}+5 x y+8 y^{2}+16 y-2=0,(-1,0) \] (a) Give the equation of the line that is tangent to the curve at the given point. \[ y= \] (b) Give the equation of the line that is normal to the curve at the given point. \[ y= \]

Step 1 ：Given the curve \(2 x^{2}+5 x y+8 y^{2}+16 y-2=0\) and the point \((-1,0)\).

Step 2 ：To find the tangent and normal lines to the curve at the given point, we first need to find the derivative of the curve. The derivative of a function gives us the slope of the tangent line at any point on the curve.

Step 3 ：Calculate the derivative of the curve, \(f' = 4x + 5y\).

Step 4 ：Substitute the given point into the derivative to find the slope of the tangent line, \(slope_{tangent} = -4\).

Step 5 ：Use the point-slope form of a line to find the equation of the tangent line, \(y = -4x - 4\).

Step 6 ：The normal line to the curve at a given point is perpendicular to the tangent line at that point. The slope of the normal line is the negative reciprocal of the slope of the tangent line.

Step 7 ：Calculate the slope of the normal line, \(slope_{normal} = 1/4\).

Step 8 ：Use the point-slope form of a line to find the equation of the normal line, \(y = x/4 + 1/4\).

Step 9 ：Final Answer: (a) The equation of the line that is tangent to the curve at the given point is \(\boxed{y = -4x - 4}\). (b) The equation of the line that is normal to the curve at the given point is \(\boxed{y = \frac{x}{4} + \frac{1}{4}}\).