Find the tangent line to the curve (x^2/16)-(y^2/9)=1 at the point (4sqrt2,3)
x/8 - (2/9)y dy/dx = 0
dy/dx = (x/8) / (2y/9) = 9x/(16y)
at (4?2,3), dy/dx = 36?2/(48) = 3?2/4
equation:
y - 3 = (3?2/4)(x - 4?2)
clean it up to whatever form is needed.
Proof:
http://www.wolframalpha.com/input/?i=plot+(x%5E2%2F16)-(y%5E2%2F9)%3D1+,+y+-+3+%3D+(3%E2%88%9A2%2F4)(x+-+4%E2%88%9A2)
Well, are you ready for some curve humor? Let's find that tangent line, but don't worry, it won't be too tangential!
First, let's rewrite the equation of the curve in a more standard form. We have (x^2/16) - (y^2/9) = 1, which can be simplified to (x^2/16) - (y^2/9) - 1 = 0. Now, to find the slope of the tangent line, we need to compute the derivative of this equation with respect to x.
Differentiating both sides with respect to x, we get (1/8)x - (2/9)y(dy/dx) = 0. Solving for dy/dx, we find dy/dx = (9/16)(x/y).
At the point (4√2, 3), the slope of the tangent line is dy/dx = (9/16)((4√2)/3). Simplifying this, we get dy/dx = (3√2)/4.
Now, we have the point (4√2, 3) and the slope dy/dx = (3√2)/4. So using the point-slope form of the equation of a line, we get y - 3 = (3√2)/4 (x - 4√2).
Therefore, the equation of the tangent line to the curve (x^2/16) - (y^2/9) = 1 at the point (4√2, 3) is y = (3√2)/4 (x - 4√2) + 3.
And that's how we keep things "tangentially" funny!
To find the equation of the tangent line to the curve at the given point, we need to find both the slope and the point.
1. Differentiate the equation with respect to x to find the slope of the tangent line:
d/dx of [(x^2/16) - (y^2/9)] = 0
(2x/16) - (2y/9) * dy/dx = 0
2. Substitute the coordinates of the given point into the equation to find the derivative, dy/dx:
(2 * 4√2)/16 - (2 * 3)/9 * dy/dx = 0
(√2/4) - (2/3) * dy/dx = 0
Solve for dy/dx:
(√2/4) = (2/3) * dy/dx
dy/dx = (√2/4) * (3/2)
= (√2/2)
3. Now that we have the slope dy/dx, we can find the equation of the tangent line using the point-slope form (y - y1) = m(x - x1), where (x1, y1) is the given point and m is the slope:
(y - 3) = (√2/2)(x - 4√2)
Simplifying,
y - 3 = (√2/2)x - 2√2
y = (√2/2)x - 2√2 + 3
y = (√2/2)x - 2√2 + 6√2/2
y = (√2/2)x + 4√2/2
Therefore, the equation of the tangent line to the curve at the point (4√2, 3) is y = (√2/2)x + 4√2/2.
To find the equation of the tangent line to a curve at a given point, you can follow these steps:
1. Differentiate the equation of the curve with respect to x.
2. Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line.
3. Use the point-slope form of a line to obtain the equation of the tangent line.
Let's work through these steps:
Step 1:
Differentiating the equation of the curve will give us the derivative function dy/dx. Let's differentiate (x^2/16) - (y^2/9) = 1 with respect to x.
Using the quotient rule for differentiation, we have:
[Derivative of (x^2/16)] - [Derivative of (y^2/9)] = 0
(Differentiating with respect to x)
(2x/16) - (2y/9) * (dy/dx) = 0
Simplifying this expression, we get:
(dx/dx) * (2x/16) - (2y/9) * (dy/dx) = 0
2x/16 - (2y/9) * (dy/dx) = 0
Step 2:
To find the slope of the tangent line, we can substitute the x-coordinate of the given point (4sqrt2) into the derivative expression.
Substituting x = 4sqrt2, we have:
2(4sqrt2)/16 - (2y/9) * (dy/dx) = 0
sqrt2/2 - (2y/9) * (dy/dx) = 0
Step 3:
Now, we have the slope of the tangent line. Next, we'll use the point-slope form of a line by substituting the given point (4sqrt2, 3) and the slope into the equation.
Using the point-slope form: y - y1 = m(x - x1)
y - 3 = (sqrt2/2)(x - 4sqrt2)
Simplifying the equation, we get the equation of the tangent line:
y = (sqrt2/2)x - 4 + 3
y = (sqrt2/2)x - 1
Therefore, the equation of the tangent line to the curve (x^2/16) - (y^2/9) = 1 at the point (4sqrt2, 3) is y = (sqrt2/2)x - 1.