If x^2 + y^2 - 2x - 4y = 20 , then an equation of the tangent line at (-3, -1) is.. (possible solutions below)

Question

a. 3x + 4y - 36 = 0
b. 4x + 3y - 35 = 0
c. 3x - 4y + 30 = 0
d. 4x - 3y - 23 = 0
e. 3x + 4y + 14 = 0
f. 4x + 3y + 15 = 0
g. 3x - 4y - 20 = 0
h. 4x - 3y + 27 = 0
i. none of these

Answer 1

In order to solve this problem, we have to use implicit differentiation.

x^2 + y^2 - 2x - 4y = 20

Now, we differentiate. Note that we're using the chain rule as we go along for y, whose derivative is dy/dx.

2x + 2y(dy/dx) - 2 - 4(dy/dx) = 0

Now, we group together the terms with dy/dx, and move everything else to the right hand side.

2y(dy/dx) - 4(dy/dx) = 2 - 2x

We can divide everything by 2, since we seem to have even coefficients.

y (dy/dx) - 2(dy/dx) = 1 - x

Now we can factor dy/dx out, getting

(dy/dx) [y - 2] = [1 - x]

And then dividing both sides by (y-2), we get

dy/dx = [1 - x] / [y - 2]

dy/dx represents the slope of the tangent line. So all we have to do is plug in (-3, -1) for x and y

dy/dx = [1 - (-3)] / [ -1 - 2]
dy/dx = [4]/[-3]

Therefore, dy/dx = -4/3

Now that we have our slope, we can use our "slope = slope" formula.

Since m = (y2 - y1)/(x2 - x1), we can substitute m = -4/3 through the points (-3,-1) and (x,y). Thus,

-4/3 = (y - (-1)) / (x - (-3))
-4/3 = (y + 1) / (x + 3)
-4(x+3) = 3(y+1)
-4x - 12 = 3y + 3
0 = 4x + 3y + 15

Therefore, the answer is (f).

Answer 2

x² + y² - 2x - 4y = 20 Differentiate with admire to x. 2x + 2y dy/dx -2 - 4 dy/dx = 0 (2y - 4)dy/dx = 2 - 2x (y - 2)dy/dx = a million - x dy/dx = (a million - xfae4153f30dfa2a04d8455e9cdd16fcfy - 2) See if the given factor is actual on the curve. (-3)² + (-a million)² - 2(-3) - 4(-a million) = 9 + a million + 6 + 4 = 20 So it relatively is. evaluate dy/dx at (-3, -a million). [a million - (-3)]/[(-a million) - 2] = -4/3 So the line has slope -4/3 and incorporates factor (-3, -a million). y + a million = (-4/3)(x + 3) 3y + 3 = -4x - 12 4x + 3y + 15 = 0

Answer 3

I don't know if the previous answer was truly a wild guess, but it was the correct one.

First, you must find the center of the circle by completing the square for x and y. You get:

(x-1)^2 + (y-2)^2 = 25

The center is at (1,2). The tangent line at (-3,-1) is the line perpendicular to the radial line connecting the center (1,2) to (-3,-1). So find the slope of the radial line:

m=(-1-2) / (-3-1) = 3/4

The slope of the tangent line must then be:

m= - 4/3

Use this in y=mx+b:

y= -(4/3)x + b

To find b, plug in one point you have for the tangent line: (-3,-1). This yields b=-5.

So, the equation of the tangent line is:

y = -(4/3)x - 5

Written in standard form, you get:

4x +3y +15=0

Thus the answer is f.

Hope this helps.

Answer 4

Find gradient of graph. Differentiate x^2 + y^2 - 2x -4y = 20 with respect to x.

You will get :
2x + (dy/dx)(2y) - 2 - 4(dy/dx) = 0

Rearranging :
(dy/dx)(2y-4) = 2-2x
dy/dx = (2-2x)/(2y-4)
= (1-x)/(y-2) note : dy/dx means gradient

Thus, at point (-3, -1) the gradient is :
dy/dx = (1-(-3))/(-1-2)
= 4/-3 = -4/3

Hence tangent at that point is :
y - (-1) = (-4/3)(x-(-3))
3y + 3 = -4x - 12
Rearrange :
4x + 3y +15 = 0

Hence answer (f)

I hope you get the idea how to do tangent questions. Its not about getting the right answer. Its about how to do it. Once you know the method, any graph under the sun is no problem for u to find the tangent. Enjoy mathematics...

Answer 5

x^2 + y^2 - 2x - 4y = 20
Centre of circle: (1,2)
Gradient of straight line passes through (1,2) and (-3,-1)
=(-1-2)/(-3-1)
=3/4
Gradient of tangent= -4/3
(y-(-1))/(x-(-3))=-4/3
3y+3=-4x-12
4x+3y+15=0

Answer: f

Answer 6

Wild guess: f