Find the equation of the tangent line and the normal line to the graph of the equation at the indicated point.
x^2-y^2=16, (5,3)
I need to show work, so answers formatted in this manner would be most appreciated. Thanks! :) :)
For the tangent line slope, you want dy/dx.
In your case you can differentiate both sides of the equation with respect to x, treating y as a function of x.
2x - 2 dy*dy/dx = 0
dy/dx = x/y = 5/3
For the equation of the line, write it in the form
(y -3)/(x-5) = slope = 5/3
y-3 = (5/3)(x-5) = 5/3 x -25/3
y = (5/3)x -16/3
Thanks so much! :)
You're a lifesaver!
To find the equation of the tangent line and the normal line to the graph of the equation at the point (5,3), we first need to find the derivative of the equation.
Given the equation x^2 - y^2 = 16, let's rearrange it to get y^2 = x^2 - 16, then take the square root on both sides, giving us y = ±√(x^2 - 16).
To find the derivative, we'll use the chain rule. The derivative of y with respect to x, or dy/dx, is obtained by differentiating the right-hand side of the equation y = ±√(x^2 - 16). Since there is a ± sign, we will consider the positive and negative roots separately.
Taking the positive root, we have y = √(x^2 - 16). Now we differentiate with respect to x, using the chain rule:
dy/dx = (1/2√(x^2 - 16)) * (2x) = x/√(x^2 - 16).
For the negative root, we have y = -√(x^2 - 16). Differentiating similarly:
dy/dx = (1/2√(x^2 - 16)) * (2x) = -x/√(x^2 - 16).
Now that we have found the derivative, the slope of the tangent line at the point (5,3) can be found by substituting x = 5 into the derivative equation:
dy/dx = x/√(x^2 - 16) = 5/√(5^2 - 16) = 5/√9 = 5/3.
Therefore, the slope of the tangent line is 5/3. To find the equation of the tangent line, we need a point on the line. We'll use the given point (5,3).
Using the point-slope form of a line, which is:
y - y₁ = m(x - x₁),
where (x₁, y₁) is a point on the line and m is the slope, we can substitute (5,3) and 5/3 into the equation:
y - 3 = (5/3)(x - 5).
Now we can simplify and rearrange to find the equation of the tangent line:
3y - 9 = 5x - 25,
3y = 5x - 16,
y = (5/3)x - 16/3.
Therefore, the equation of the tangent line to the graph of the equation x^2 - y^2 = 16 at the point (5,3) is y = (5/3)x - 16/3.
To find the equation of the normal line, we first need to find the negative reciprocal of the slope of the tangent line, which is -3/5. Following the same process as above, substituting (5,3) and -3/5 into the point-slope form equation:
y - 3 = (-3/5)(x - 5),
simplifying and rearranging:
5y - 15 = -3x + 15,
5y = -3x + 30,
y = (-3/5)x + 6.
Hence, the equation of the normal line to the graph of the equation x^2 - y^2 = 16 at the point (5,3) is y = (-3/5)x + 6.