Find an equation of the tangent line to the graph of x^2+3y^2=4 at the point (1,1)
A. y+1=-1/3(x+1)
B. y-1=-x/3y(x-1)
C. x+3y=2
D. y-1=-1/3(x-1)
E. None of these
dy/dx = slope = m
for this ellipse
2 x dx/dx + 6 y dy/dx = 0
at(1,1)
2(1)+6(1)dy/dx = 0
dy/dx = m = -1/3
so
y = -(1/3) x + b
at (1,1)
1 = -1/3) + b
b = 4/3
so
y = -(1/3) x + 4/3
or
3 y - -x + 4
E. None of these
To find the equation of the tangent line to the graph of x^2 + 3y^2 = 4 at the point (1,1), we need to find the slope of the tangent line and then use the point-slope form of a linear equation.
Step 1: Differentiate the equation with respect to x to find the derivative.
Differentiating x^2 + 3y^2 = 4 with respect to x, we get:
2x + 6y(dy/dx) = 0
Step 2: Solve for dy/dx.
6y(dy/dx) = -2x
dy/dx = -2x / 6y
Simplifying further, dy/dx = -x / 3y
Step 3: Substitute the coordinates of the point (1,1) into the slope equation.
At the point (1,1), x = 1 and y = 1. Substituting these values, we get:
dy/dx = -1 / 3
Step 4: Use the point-slope form of a linear equation.
Using the slope (-1/3) and the point (1,1) in the point-slope form of a linear equation:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is the point,
we have:
y - 1 = (-1/3)(x - 1)
Simplifying, we get:
y - 1 = -1/3x + 1/3
Adding 1 to both sides, we obtain:
y = -1/3x + 4/3
So, the equation of the tangent line to the graph of x^2 + 3y^2 = 4 at the point (1,1) is:
y - 1 = -1/3(x - 1)
Therefore, the correct answer is A. y + 1 = -1/3(x + 1).
To find the equation of the tangent line to the graph of x^2 + 3y^2 = 4 at the point (1,1), we'll need to use calculus. Here are the steps you can follow to find the equation:
Step 1: Differentiate the equation with respect to x.
Taking the derivative of both sides of the equation will give us the slope of the tangent line at any given point.
Differentiating x^2 + 3y^2 = 4 with respect to x, we get:
2x + 6y * dy/dx = 0
Step 2: Solve for dy/dx.
Rearranging the equation, we have:
dy/dx = -2x / (6y)
Step 3: Substitute the coordinates of the given point.
Substituting the coordinates (1,1) into the equation in step 2, we have:
dy/dx = -2(1) / (6(1))
dy/dx = -2 / 6
dy/dx = -1/3
Step 4: Write the equation of the tangent line using the point-slope form.
Using the slope (-1/3) and the point (1,1), we can write the equation of the tangent line in the point-slope form:
y - y1 = m(x - x1)
Substitute y1 = 1, x1 = 1, and m = -1/3, we get:
y - 1 = (-1/3)(x - 1)
y - 1 = (-1/3)x + 1/3
y = (-1/3)x + 4/3
Therefore, the equation of the tangent line to the graph of x^2 + 3y^2 = 4 at the point (1,1) is y = (-1/3)x + 4/3.
Comparing this equation with the given options, we can see that it matches option D. So the correct answer is D.