use implicit differentiation to find an equation of the tangent line to the curve at the given point.
9x^2+xy+9y^2=19, (1,1) (ellipse)
18x + y + xy' + 18yy' = 0
y' = -(18x+y)/(x+18y)
at (1,1) y' = -1
so, the line is
y-1 = -(x-1)
To find the equation of the tangent line to the given curve at the point (1,1), we will use implicit differentiation. The given equation is 9x^2 + xy + 9y^2 = 19 (ellipse).
Step 1: Differentiate both sides of the equation with respect to x.
d/dx(9x^2 + xy + 9y^2) = d/dx(19)
Step 2: Apply the product rule and chain rule as necessary.
18x + 1y + x(dy/dx) + 18y(dy/dx) = 0
Step 3: Simplify the equation.
(1 + x(dy/dx))y + 18x + 18y(dy/dx) = 0
Step 4: Substitute the coordinates of the given point (1,1) into the equation.
(1 + 1(dy/dx))(1) + 18(1) + 18(1)(dy/dx) = 0
Step 5: Solve for dy/dx.
1 + dy/dx + 18 + 18(dy/dx) = 0
19 + 19(dy/dx) = 0
19(dy/dx) = -19
dy/dx = -1
Therefore, the slope of the tangent line at the point (1,1) is -1.
Step 6: Use the slope-intercept form of a line (y = mx + b) to find the equation of the tangent line.
Using the point-slope form of a line, we know that the equation of the tangent line is:
y - 1 = -1(x - 1)
Simplifying the equation further:
y - 1 = -x + 1
y = -x + 2
Therefore, the equation of the tangent line to the given ellipse at the point (1,1) is y = -x + 2.
To find the equation of the tangent line to the ellipse 9x^2 + xy + 9y^2 = 19 at the given point (1, 1), we can use implicit differentiation and the point-slope form of a line.
Implicit differentiation involves differentiating both sides of the equation with respect to x, treating y as a function of x.
Let's start by differentiating the given equation with respect to x term by term:
d/dx (9x^2 + xy + 9y^2) = d/dx (19)
To differentiate each term, we use the rules of differentiation:
d/dx (9x^2) = 18x
d/dx (xy) = y + x(dy/dx)
d/dx (9y^2) = 18y(dy/dx)
The derivative of the constant term 19 is 0.
Now, let's substitute the coordinates of the given point (1, 1) into the equation and solve for dy/dx.
Substituting x = 1 and y = 1 into the equation, we get:
18(1) + 1(1) + 18(1)(dy/dx) = 0
18 + 1 + 18(dy/dx) = 0
19 + 18(dy/dx) = 0
18(dy/dx) = -19
dy/dx = -19/18
Now, we have the value of dy/dx at the point (1, 1). The slope of the tangent line is given by dy/dx.
Using the point-slope form of a line, we have the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values, we get:
y - 1 = (-19/18)(x - 1)
Simplifying and rearranging, we have:
y = (-19/18)x + 37/18
Therefore, the equation of the tangent line to the ellipse 9x^2 + xy + 9y^2 = 19 at the point (1, 1) is y = (-19/18)x + 37/18.