Given the equation of a closed curve is x^2-2x+4y^2+16y+1=0.Find the coordinates of the two points where the lines tangent to the curve are vertical.
2x - 2 + 8y dy/dx + 16dy/dx = 0
dy/dx(8y + 16) = 2 - 2x
dy/dx = (2-2x)/(8y + 16)
if the tangents are vertical, then their slopes are undefined, that is, the denominator of dy/dx = 0
8y + 16 = 0
y = -2
sub into original equation:
x^2 - 2x + 8 - 32 + 1 = 0
x^2 - 2x - 23 = 0
x = (2 ± √96)/2
= 1 ± 2√6
the points are (1+2√6 , -2) and (1 - 2√6 , -2)
This can be easily confirmed by noting that the equation is an ellipse:
(x-1)^2/16 + (y+2)^2/4 = 1
The major axis is on the line y = -2, so that's where the tangents are vertical.
To find the coordinates of the points where the tangent lines are vertical, we need to find the points on the curve where the derivative with respect to x is undefined. This is because the slope of a vertical line is undefined.
Let's solve this step by step:
Step 1: Rewrite the equation in the standard form of a conic section.
We are given the equation:
x^2 - 2x + 4y^2 + 16y + 1 = 0
Completing the square for the x terms:
(x^2 - 2x) + 4y^2 + 16y + 1 = 0
(x^2 - 2x + 1) + 4y^2 + 16y + 1 = 1
(x - 1)^2 + 4y^2 + 16y + 2 = 1
Completing the square for the y terms:
(x - 1)^2 + 4(y^2 + 4y) + 2 = 1
(x - 1)^2 + 4(y^2 + 4y + 4) + 2 = 1 + 4(4)
(x - 1)^2 + 4(y + 2)^2 + 2 = 17
So, the equation of the curve in standard form is:
(x - 1)^2 + 4(y + 2)^2 = 15
Step 2: Find the derivative dy/dx.
To find the derivative, we need to differentiate both sides of the equation with respect to x.
(x - 1)^2 + 4(y + 2)^2 = 15
Differentiating, we get:
2(x - 1) + 4(2)(y + 2) * dy/dx = 0
2(x - 1) + 8(y + 2) * dy/dx = 0
2(x - 1) + 8(y + 2)dy/dx = 0
Step 3: Solve for dy/dx.
To find the values of x when the tangent lines are vertical, dy/dx must be undefined.
Since the slope is undefined when the denominator is zero, we know that dy/dx will be undefined when the coefficient of dy/dx (8(y + 2)) is zero.
Setting 8(y + 2) = 0:
8(y + 2) = 0
y + 2 = 0
y = -2
Step 4: Find the corresponding x-values.
Now that we have found the value of y when dy/dx is undefined, we can substitute this value into the original equation:
(x - 1)^2 + 4(y + 2)^2 = 15
Substituting y = -2:
(x - 1)^2 + 4(-2 + 2)^2 = 15
(x - 1)^2 = 15
We have reduced the equation to a quadratic equation in x. Solving this equation will give us the corresponding x-values.
Solving, we find:
x - 1 = ±√15
x = 1 ± √15
Therefore, the coordinates of the two points where the tangent lines are vertical are:
(1 + √15, -2) and (1 - √15, -2).