The parametric equations of a curve are x = 4t and y = 4 − t2.
Find the equations of the normals to the curve at the points where
the curve meets the x-axis. Hence, find the point of intersection of these normals.
I bet you, that second equation is y = 4-t^2
form the first: x^2 = 16t^2 --> t^2 = x^2/16
from the 2nd: t^2 = 4 - y
so x^2/16 = 4-y
x^2 = 64 - 16y , which is a parabola opening downwards.
derivative: 2x = -16 dy/dx
dy/dx = -x/8
x-intercepts of the parabola, let y = 0
x = ±8
so we have intercepts at (8,0) and (-8,0)
at (8,0), slope of tangent = -8/8 = -1
so the normal has a slope of +1
equation of normal: y = 1(x-8)
at (-8,0), slope of tangent = 1
so slope of normal = -1
equation of normal : y = -(x+8)
solving these two normals:
x-8 = -x-8
2x = 0
x = 0 , obvious from the symmetry
then y = -8
the normals intersect at (0,-8)
To find the equations of the normals to the curve at the points where it meets the x-axis, we first need to determine the points of intersection.
Since the curve meets the x-axis when y = 0, we can substitute y = 0 into the second equation:
0 = 4 - t^2
Rearrange the equation to solve for t:
t^2 = 4
Taking the square root of both sides, we get:
t = ±2
So, the curve intersects the x-axis at two points: when t = 2 and when t = -2.
To find the corresponding points on the curve, we can substitute these values of t into the given parametric equations:
For t = 2:
x = 4(2) = 8
y = 4 - (2^2) = 0
So, one point of intersection is (8, 0).
For t = -2:
x = 4(-2) = -8
y = 4 - ((-2)^2) = 0
So, the other point of intersection is (-8, 0).
Now, to find the equations of the normals at these points, we need to find the slopes of the tangents at these points.
The slope of the tangent at any point on the curve can be found by taking the derivative of y with respect to x:
dy/dx = (dy/dt) / (dx/dt)
Given x = 4t and y = 4 - t^2, we can calculate the derivatives:
dx/dt = 4
dy/dt = -2t
Substituting these values into the slope formula, we get:
dy/dx = (-2t) / 4
dy/dx = -t/2
So, the slope of the tangent at any point on the curve is -t/2.
To find the slope of the normal, we take the negative reciprocal of the tangent's slope:
m_normal = -1 / (-t/2)
m_normal = 2/t
Therefore, the slopes of the normals at the points of intersection are 2/2 = 1 for t = 2 and 2/-2 = -1 for t = -2.
Using the point-slope form of a line (y - y₁ = m(x - x₁)), we can write the equations of the normals:
For the point (8, 0) with slope 1:
y - 0 = 1(x - 8)
y = x - 8
For the point (-8, 0) with slope -1:
y - 0 = -1(x + 8)
y = -x - 8
Now, to find the point of intersection of these normals, we can solve the system of equations:
y = x - 8
y = -x - 8
One way to solve this system is by adding the two equations:
(x - 8) + (-x - 8) = 0
-16 = 0
This equation is not true, so there is no point of intersection between the two normals.
To find the equations of the normals to the curve at the points where it meets the x-axis, we need to find the points of intersection first.
The curve meets the x-axis when y = 0. Therefore, substitute y with 0 in the given parametric equation:
0 = 4 - t²
Rearrange the equation to solve for t:
t² = 4
Taking the square root of both sides:
t = ± 2
So, there are two points where the curve meets the x-axis: when t = 2 and when t = -2.
Now, we need to find the corresponding x-coordinates for these values of t. Plugging t = 2 and t = -2 into the given parametric equation for x:
For t = 2:
x = 4(2) = 8
For t = -2:
x = 4(-2) = -8
Now we have the two points of intersection with the x-axis: (8, 0) and (-8, 0).
To find the equations of the normals at these points, we need the gradients (or slopes) of the tangents to the curve. The gradient of a tangent to a parametric curve is given by dy/dx.
Differentiating both x and y with respect to t:
dx/dt = 4
dy/dt = -2t
To find dy/dx, divide dy/dt by dx/dt:
dy/dx = (-2t) / 4 = -t/2
Now we have the gradient of the tangent to the curve. To find the gradient of the normal at each point, we take the negative reciprocal of the gradient of the tangent. This means flipping the fraction and changing its sign.
For t = 2:
Gradient of the tangent, m = -t/2 = -(2)/2 = -1
Gradient of the normal, m' = 1
For t = -2:
Gradient of the tangent, m = -t/2 = -(-2)/2 = 1
Gradient of the normal, m' = -1
Now we can find the equation of the normal at each point using the point-gradient form of a straight line equation:
For the point (8, 0) with gradient m' = 1:
(y - y₁) = m'(x - x₁)
(y - 0) = 1(x - 8)
y = x - 8
For the point (-8, 0) with gradient m' = -1:
(y - y₁) = m'(x - x₁)
(y - 0) = -1(x - (-8))
y = -x - 8
Therefore, the equations of the normals to the curve at the points where it meets the x-axis are y = x - 8 and y = -x - 8.
Finally, to find the point of intersection of these normals, we can solve the two equations together. By setting the right sides of the equations equal to each other:
x - 8 = -x - 8
Adding x to both sides:
2x - 8 = -8
Adding 8 to both sides:
2x = 0
Dividing both sides by 2:
x = 0
Now substituting x = 0 into either equation, we find:
y = 0 - 8 = -8
So the point of intersection of the normals is (0, -8).