how to find the slope of the tangent line to each curve at given point? see the details plzzz?

Question

i have to questions and both of them is about to find the slope of the tangent but with different point.

question nr. 1 : find the slope of the tangent line to each curve at the given point. x^2-y^2=16; (5, 3)

question nr. 2: find the slope of the tangent line to each curve at the given point: y^2+3x-8y+3=0; (4, 3)

plzzzzzzzzzzzz help me and wirte it with step by step how to do it plzzzzzzzzzzzzz .

Answer 1

that is implicit differential
problem 1
x^2 -y^2 = 16
take derivative both sides which respect to x
2x -2ydy/dx =0
2x= 2ydy/dx
2x/2y = dy/dx

x/y = dy/dx
put (5,3)
then slope dy/dx = 5/3


problem 2
y^2 +3x-8y+3=0
samething takes derivative both sides respected to x
2ydy/dx +3 -8dy/dx =0
dy/dx(2y-8) +3 =0
dy/dx = -3/{2y-8}
slope dy/dx = -3/{2*3-8} = -3/-2 =3/2
hope it helps, answer can be found be different ways that depends on what levels are you in now

Answer 2

The other methods already above are correct, but here is one more way.

1. Rearrange the equation so y is by itself;
y=(x^2-16)^(1/2)
Take the derivative of this equation,
f(x)=x/((x^2-16)^(1/2))
Then plug in the value for x
f(5)= 5/3 ***Answer

2. Rearrange to solve for x, because it makes it a lot easier
x=-y^2/2 + 8/3y -1
Take the derivative of x with respect to y
f(y) = f(3) = -2(3)/3 + 8/3 = 2/3 ***slope
With this method, you must keep in mind that my slope is now in the form dx/dy = 2/3, when most of the time you will need to find dy/dx,so in this case, the most important part is to flip your answer so you get dy/dx = 3/2. Hope this method helps you! = )

Answer 3

1) Well, assume that the slope of this line is k, the line equation is:
y - 3 = k(x - 5), or y = kx + (3 - 5k)
The intersection point should be the solution to the following equation:
x^2 - {kx + (3 - 5k)}^2 = 16
or (1 - k^2)x^2 - 2k(3 - 5k)x - (3 - 5k)^2 - 16 = 0
Since this line is tangent to x^2 - y^2 = 16, they only have one intersection point, and hence the discriminant of this quadratic should be zero:
0 = b^2 - 4ac = 4k^2(3 - 5k)^2 + 4(1 - k^2)[(3 - 5k)^2 +16]
= 36k^2 - 120k + 100
= 4(3k - 5)^2
Therefore k = 5/3
(2) please follow the same way, and you can solve this problem readily. Good Luck!

Answer 4

the 1st answer (above) isn't maximum suitable. you are able to desire to honestly discover the slope as dy/dx (not dy/dt and dx/dt because of the fact the 1st guy or woman used). i'm assuming you're in a calculus class.... dx = 6t^2 dt and dy = (8 + 2t) dt slope = dy/dx dy/dx = [ (8 + 2t)dt ] / [ 6t^2 dt ] yet extremely we choose the slope to be a million: a million = (8 + 2t) / (6t^2) 6t^2 = 8 + 2t 6t^2 - 2t - 8 = 0 3t^2 - t - 4 = 0 t = -a million, t = 4/3 now plugging -a million and four/3 into the x & y equations: t = -a million: x = 2t^3 = -2 y = 3 + 8t + y^2 = -4 (-2, -4) t = 4/3: x = 2t^3 = 128/27 y = 3 + 8t + y^2 = 139/9 (128/27, 139/9)