Find the slope of the tangent line to the ellipse
(x^2)/4 + (y^2)/9 = 1 at the point (x,y)
I would first multiply by 36 to get rid of the nasties.
9x^2 + 4y^2 = 36
18x + 8y dy/dx = 0
dy/dx = -18x/(8y) = -9x/(4y)
all done.
Well, finding the slope of the tangent line to an ellipse may seem daunting, but fear not! I'm here to make it as fun as a circus act!
To find the slope of the tangent line at the point (x, y) on the ellipse (x^2)/4 + (y^2)/9 = 1, we first need to find the derivative of the equation. So let's get this equation in tip-top shape!
Rearranging the equation a bit, we have:
(x^2)/4 = 1 - (y^2)/9
If we differentiate both sides with respect to x, we get:
(1/4)*(2x) = - (2y/9) * (dy/dx)
Simplifying that, we have:
x/2 = - (y/9) * (dy/dx)
Now, solving for dy/dx, we get:
dy/dx = - (9x)/(2y)
And there we have it, folks! The slope of the tangent line to the ellipse (x^2)/4 + (y^2)/9 = 1 at the point (x, y) is - (9x)/(2y).
Remember, math can be as amusing as a clown performance if you approach it with a smile!
To find the slope of the tangent line to the ellipse, we can use differentiation.
First, let's differentiate both sides of the equation with respect to x:
d/dx [(x^2)/4 + (y^2)/9] = d/dx 1
Now, we'll differentiate each term separately:
(d/dx) (x^2)/4 + (d/dx) (y^2)/9 = 0
To differentiate (x^2)/4, we can use the power rule, which states that for any function f(x) = x^n, the derivative is given by f'(x) = n*x^(n-1). Applying the power rule, we have:
(1/4) * d/dx (x^2) = (1/4) * (2x) = (1/2)x
To differentiate (y^2)/9 with respect to x, we need to apply the chain rule. The chain rule states that for any function f(g(x)), the derivative is given by f'(g(x)) * g'(x). In this case, g(x) = y(x), so we can write y^2 = (3y)^2 = 9y^2. Applying the chain rule, we have:
(1/9) * 9 * d/dx y^2 = 2y * (dy/dx) = 2y * dy/dx
Substituting this back into our equation:
(1/2)x + (2y * dy/dx) = 0
Now, let's solve for dy/dx:
2y * dy/dx = -(1/2)x
Dividing both sides by 2y:
dy/dx = -(1/4)(x/y)
Now, to find the slope of the tangent line at the point (x, y), we substitute the values into the equation we obtained:
Slope = dy/dx evaluated at (x, y) = -(1/4)(x/y)
So, the slope of the tangent line to the ellipse at any point (x, y) is -(1/4)(x/y).
To find the slope of the tangent line to the ellipse at the point (x, y), we can use the derivative of the equation of the ellipse.
Step 1: Write the equation of the ellipse in general form:
(x^2)/a^2 + (y^2)/b^2 = 1
For this particular ellipse, a = 2 and b = 3, so the equation becomes:
(x^2)/4 + (y^2)/9 = 1
Step 2: Differentiate both sides of the equation with respect to x. Differencing the equation, we get:
(2x)/4 + (2y)(dy/dx)/9 = 0
Simplifying this further, we have:
dy/dx = -((2x)/(2y))*((4)/(9))
dy/dx = -x/(2y)*(4/9)
dy/dx = -2x/(9y)
Step 3: Substitute the coordinates of the point (x, y) into the derivative equation to find the slope at that point.
For the point (x, y), the slope of the tangent line is:
slope = -2x/(9y)
Simply substitute the given values of x and y into the equation above, and you will find the value of the slope at the point (x, y) on the ellipse.