the normal to a parabola y^2=4ax at the point t,where t is not 0 meets the curve again at the point t'.find t' in terms of t.determine a point on the x-axis where the tangent at t' meets the x-axis.
To find the normal to the parabola and the point t', we need to first find the slope of the tangent at point t on the parabola.
Given that the equation of the parabola is y^2 = 4ax, we can differentiate it implicitly with respect to x to get:
2y * dy/dx = 4a
Now, let's find dy/dx at the point t. We substitute y = t into the equation above:
2t * dy/dx = 4a
Therefore, dy/dx = 2a/t.
The slope of the normal to the parabola at point t will be the negative reciprocal of the slope of the tangent. So, the slope of the normal is:
m = -1 / (2a/t) = -t / (2a)
Now, let's find the equation of the normal to the parabola using the point-slope form:
(y - y1) = m(x - x1)
Since the normal passes through (t, t^2/4a), we substitute these coordinates into the equation above:
(y - t^2/4a) = (-t / (2a))(x - t)
Now, let's find the coordinates of the point t' where the normal intersects the parabola again.
Substitute the equation of the normal into the equation of the parabola and solve for x:
(y - t^2/4a)^2 = 4ax
[( -t / (2a))(x - t) - t^2/4a]^2 = 4ax
Expanding and simplifying, we get:
(x - t)^2 = -t^2
Taking square root on both sides, we get:
x - t = ±i * t
Simplifying further, we have:
x = t ± it
Therefore, the point t' where the normal intersects the parabola again has coordinates (t ± it, -(t^2/4a)).
Now, let's find a point on the x-axis where the tangent at t' intersects the x-axis.
To find the slope of the tangent at t', we differentiate the equation of the parabola implicitly with respect to x:
2y * dy/dx = 4a
When y = -(t^2/4a), we substitute these values into the equation above:
2 * (-(t^2/4a)) * dy/dx = 4a
Simplifying further, we get:
-(t^2/2) * dy/dx = 4a
Therefore, dy/dx = -8a / t^2.
The slope of the tangent at t' will be the negative reciprocal of dy/dx. So, the slope of the tangent is:
m = -1 / (-(8a/t^2)) = t^2 / (8a)
The equation of the tangent to the parabola at t' will be:
(y - y1) = m(x - x1)
Since the tangent passes through (t ± it, -(t^2/4a)), we substitute these coordinates into the equation above:
(y + t^2/4a) = (t^2 / (8a))(x - t ± it)
To find the point on the x-axis where the tangent intersects, we substitute y = 0 into the equation above and solve for x:
0 + t^2/4a = (t^2 / (8a))(x - t ± it)
Simplifying, we get:
t^2/4a = (t^2 / (8a))(x - t ± it)
Multiplying both sides by (8a / t^2):
2 = 2(x - t ± it)
Therefore, x - t ± it = 1
For the point of intersection with the x-axis, y = 0. So, substituting y = 0 into the equation of the tangent, we get:
0 + t^2/4a = (t^2 / (8a))(x - t ± it)
Simplifying, we get:
t^2/4a = (t^2 / (8a))(x - t ± it)
Since y = 0, we have:
t^2/4a = (t^2 / (8a))(x - t ± it)
Multiplying both sides by (8a / t^2):
2 = 2(x - t ± it)
Therefore, x - t ± it = 1
Simplifying further, we get:
x = t ± it + 1
Therefore, a point on the x-axis where the tangent at t' intersects the x-axis is (t ± it + 1, 0).
In summary:
- The point t' where the normal intersects the parabola again has coordinates (t ± it, -(t^2/4a)).
- A point on the x-axis where the tangent at t' intersects the x-axis is (t ± it + 1, 0).