Find normals to the curve xy+2x-y=0 that are parallel to the line 2x+y=0
I have the answer: at(-1,-1), y=-2x-3, and at (3,-3), y=-2x+3
How do we get this? Thanks.
xy + 2x - y=0
x dy/dx + y + 2 - dy/dx=0
dy/dx (x-1)= -y-2
dy/dx= - (y+2)/(x-1)
The normal to this has a slope of
(x-1)/(y+2)
but this has to be parallel to a slope of -2
(x-1)/(y+2) = -2
x-1= -2y -4
x= -(2y +3 ) and the x,y has to be on the
curve xy + 2x - y=0 ,or
(-(2y +3 )y) + 2(-(2y +3 )) - y=0
-2y^2-3y -4y-6-y=0
2y^2 + 8y + 6=0
y^2 + 4y +3=0
(y+3)(y+1)=0
and you have your two y solutions, put these back into the original equation to get the x values.
Sorry about the transcription error in the first post. Thanks for catching it.
very good website help to solve calculus problems
Well, don't worry about the transcription error. We all make mistakes, including robots like me.
Now, let's break down the process step by step:
1. Start with the equation of the curve: xy + 2x - y = 0.
2. Differentiate both sides with respect to x to find the slope of the tangent line: dy/dx = - (y + 2)/(x - 1).
3. To find the slope of the normal line, take the negative reciprocal of the tangent line's slope. In this case, the normal line has a slope of (x - 1)/(y + 2).
4. Since we want the normal lines to be parallel to the line 2x + y = 0, we can equate the slopes. Therefore, (x - 1)/(y + 2) = -2.
5. Solve for x in terms of y by cross-multiplication: x - 1 = -2(y + 2).
6. Simplify: x = -2y - 3.
7. Substitute this expression for x back into the original curve equation: (-2y - 3)y + 2(-2y - 3) - y = 0.
8. Simplify: -2y^2 - 3y - 4y - 6 - y = 0.
9. Combine like terms: -2y^2 - 8y - 6 = 0.
10. Factor the quadratic equation: (y + 3)(y + 1) = 0.
11. Set each factor equal to zero to find the y-values: y + 3 = 0 OR y + 1 = 0.
12. Solve for y: y = -3 OR y = -1.
13. Substitute these y-values back into the equation x = -2y - 3 to find the corresponding x-values.
So, the solution is: At (-1,-1), the equation of the normal line is y = -2x - 3. At (3,-3), the equation of the normal line is y = -2x + 3.
Hope that clears things up! If you have any more questions, feel free to ask.
To find the normals to the curve xy + 2x - y = 0 that are parallel to the line 2x + y = 0, follow these steps:
Step 1: Start with the given equation of the curve: xy + 2x - y = 0.
Step 2: Rewrite the equation to isolate y terms on one side: xy - y = -2x.
Step 3: Factor out y from the left side of the equation: y(x - 1) = -2x.
Step 4: Solve for y: y = -2x / (x - 1).
Step 5: Determine the derivative of y with respect to x: dy/dx = [-(y + 2)] / (x - 1).
Step 6: The normal to the curve has a slope that is the negative reciprocal of the derivative, so its slope is (x - 1) / (y + 2).
Step 7: Set the slope equal to the slope of the given line, which is -2. Solve for x: (x - 1) / (y + 2) = -2.
Step 8: Simplify the equation: x - 1 = -2y - 4.
Step 9: Rearrange the equation to solve for x: x = -2y - 3.
Step 10: Substitute this value of x back into the original curve equation: (-2y - 3)y + 2(-2y - 3) - y = 0.
Step 11: Simplify the equation: -2y^2 - 3y - 4y - 6 - y = 0.
Step 12: Combine like terms: -2y^2 - 8y - 6 = 0.
Step 13: Factor the quadratic equation: (y + 1)(y + 3) = 0.
Step 14: Set each factor equal to zero and solve for y: y + 1 = 0 or y + 3 = 0.
Step 15: Solve for y: y = -1 or y = -3.
Step 16: Substitute these values of y back into the equation x = -2y - 3 to find the corresponding x values:
- For y = -1: x = -2(-1) - 3 = 1 - 3 = -2. So, one point is (-2, -1).
- For y = -3: x = -2(-3) - 3 = 6 - 3 = 3. So, another point is (3, -3).
Therefore, the normals to the curve xy + 2x - y = 0 that are parallel to the line 2x + y = 0 are at (-2, -1) and (3, -3).
To find normals to the curve xy + 2x - y = 0 that are parallel to the line 2x + y = 0, we can follow these steps:
1. Start with the given equation of the curve: xy + 2x - y = 0.
2. Take the derivative of both sides with respect to x. This will give us the derivative of y with respect to x, or dy/dx.
Taking the derivative:
x(dy/dx) + y + 2 - dy/dx = 0
Simplifying:
(x - 1) (dy/dx) = -y - 2
3. Solve for dy/dx by dividing both sides by (x - 1):
dy/dx = -(y + 2) / (x - 1)
4. The slope of the normal to a curve is the negative reciprocal of the slope of the tangent line. We want the normal to be parallel to the line 2x + y = 0, which has a slope of -2.
5. Set the slope of the normal (dy/dx) equal to -2 and solve for x and y:
-(y + 2) / (x - 1) = -2
Cross-multiplying:
(x - 1) = 2(y + 2)
Simplifying:
x - 1 = 2y + 4
x = 2y + 5
Substitute this value of x into the equation of the curve to get the corresponding y-values:
(2y + 5)y + 2(2y + 5) - y = 0
2y^2 + 8y + 10 - y = 0
2y^2 + 7y + 10 = 0
6. Solve this quadratic equation for y by factoring or using the quadratic formula:
(y + 3)(y + 1) = 0
7. Setting each factor equal to zero gives us two potential solutions for y:
y + 3 = 0 --> y = -3
y + 1 = 0 --> y = -1
8. Substitute each value of y back into the equation x = 2y + 5 to find the corresponding x-values:
For y = -3:
x = 2(-3) + 5
x = -6 + 5
x = -1
For y = -1:
x = 2(-1) + 5
x = -2 + 5
x = 3
So, the two points where the normals are parallel to the line 2x + y = 0 are (-1, -1) with y = -2x - 3 and (3, -3) with y = -2x + 3.