Given that x3-axy+3ay2=3a3 Find the equations for the tangent and normal lines at point (a,a)
To find the equations of the tangent and normal lines at the point (a, a), we need to determine the slope of the curve at that point.
1. Step 1: Differentiate the equation with respect to x to find the slope of the curve.
Taking the derivative of x^3 - axy + 3ay^2 = 3a^3 with respect to x, we get:
3x^2 - a(dy/dx) - ay(dy/dx) + 6ay(dy/dx) = 0
2. Step 2: Substituting the coordinates of the point (a, a) into the equation.
Plugging in x = a and y = a into the derivative equation, we get:
3a^2 - a(dy/dx) - a^2(dy/dx) + 6a^2(dy/dx) = 0
Simplifying, we have:
3a^2 - a(dy/dx) - a^2(dy/dx) + 6a^2(dy/dx) = 0
Combining like terms, we obtain:
(3a^2 - a^2) - a(dy/dx) + 6a^2(dy/dx) = 0
Factoring out dy/dx, we get:
(2a^2) - a(dy/dx) + 6a^2(dy/dx) = 0
Simplifying further:
2a^2 + (5a^2 - a)(dy/dx) = 0
(7a^2 - a)(dy/dx) = -2a^2
3. Step 3: Solve for dy/dx.
Dividing both sides by (7a^2 - a), we have:
dy/dx = -2a^2 / (7a^2 - a)
Simplifying the expression, we get:
dy/dx = -2a^2 / (a(7a - 1))
dy/dx = -2a / (7a - 1)
4. Step 4: Find the slope of the tangent line.
At the point (a, a), the slope of the tangent line is equal to the derivative evaluated at that point. Substituting x = a and y = a into the equation, we have:
dy/dx = -2a / (7a - 1)
Therefore, the slope of the tangent line is -2a / (7a - 1).
5. Step 5: Find the slope of the normal line.
The normal line is perpendicular to the tangent line, so the slope of the normal line is the negative reciprocal of the slope of the tangent line.
Thus, the slope of the normal line is given by:
-1 / (-2a / (7a - 1))
Simplifying, we get:
(7a - 1) / 2a
6. Step 6: Write the equations for the tangent and normal lines.
We now have the slope and a point (a, a) on the lines. By using the point-slope form of a linear equation, the equations for the tangent and normal lines can be written as follows:
Tangent line: y - a = ( -2a / (7a - 1) ) * (x - a)
Normal line: y - a = ( (7a - 1) / 2a ) * (x - a)
Simplifying these equations further is possible, but this is the general form for the tangent and normal lines at the point (a, a).
This is how you can find the equations of the tangent and normal lines at the point (a, a) for the given curve x^3 - axy + 3ay^2 = 3a^3.