Find the slope of the tangent line to the ellipse x^2/4 + y^2/16= 1 at the point (x,y)
use implicit differentiation:
x/2 + y/8 y' = 0
y' = -4x/y
To find the slope of the tangent line to the ellipse x^2/4 + y^2/16 = 1 at the point (x, y), we need to find the derivative of the equation with respect to x.
Starting with the equation of the ellipse:
x^2/4 + y^2/16 = 1
Let's rewrite it to isolate y:
y^2/16 = 1 - x^2/4
Multiplying both sides by 16:
y^2 = 16 - 4x^2
Taking the square root of both sides:
y = ±√(16 - 4x^2)
Since we want to find the slope of the tangent line, we need to find the derivative dy/dx. Let's take the derivative of both sides of the equation:
dy/dx = ± d(√(16 - 4x^2))/dx
Using the chain rule, we have:
dy/dx = ± (1/2)*d(16 - 4x^2)^(-1/2)/dx
Applying the power rule, we get:
dy/dx = ± (1/2) * (-4x^2)' * (16 - 4x^2)^(-3/2)
dy/dx = ± (-2x) / (2 * √(16 - 4x^2))
Simplifying further:
dy/dx = ± (-x) / √(16 - 4x^2)
Thus, the slope of the tangent line to the ellipse at the point (x, y) is given by ± (-x) / √(16 - 4x^2).
To find the slope of the tangent line to the ellipse at the point (x, y), we need to find the derivative of the ellipse equation with respect to x and evaluate it at the given point.
The equation of the ellipse is:
x^2/4 + y^2/16 = 1
To find the derivative with respect to x, we need to differentiate each term of the equation with respect to x. Let's differentiate both sides:
d/dx (x^2/4) + d/dx (y^2/16) = d/dx (1)
Now let's evaluate each term:
(1/4) * d/dx (x^2) + (1/16) * d/dx (y^2) = 0
To differentiate x^2 with respect to x, we use the power rule, which states that d/dx (x^n) = n * x^(n-1). Applying this rule, we get:
(1/4) * 2x + (1/16) * d/dx (y^2) = 0
Simplifying the equation:
(1/2) * x + (1/16) * d/dx (y^2) = 0
Now, let's differentiate y^2 with respect to x. Since y is a function of x, we need to use the chain rule. The chain rule states that d/dx (f(g(x))) = f'(g(x)) * g'(x), where f'(x) is the derivative of f(x) and g'(x) is the derivative of g(x).
Let y be a function of x, y = f(x), then y^2 = f(x)^2. Applying the chain rule, we get:
d/dx (y^2) = d/dx (f(x)^2) = 2f(x) * f'(x)
Substituting this back into the equation:
(1/2) * x + (1/16) * 2y * dy/dx = 0
Since we want to find the slope at the point (x, y), we can substitute the values of x and y into the equation. After substituting, if the equation is solvable, we can solve it to find dy/dx, which represents the slope of the tangent line at the point (x, y).