The tangents to a parabola with focus (1,2) are x+y=0 and x-y=o. Find the length of latus rectum.
To find the length of the latus rectum of a parabola, we need to determine the equation of the parabola.
Given that the tangents to the parabola are x + y = 0 and x - y = 0, we can write their equations in slope-intercept form (y = mx + c) to get the slopes of the tangents.
For the tangent x + y = 0:
Rewriting it in slope-intercept form, we have: y = -x + 0
Comparing this equation with y = mx + c, we can see that the slope (m) is -1.
For the tangent x - y = 0:
Rewriting it in slope-intercept form, we have: y = x + 0
Again, comparing with y = mx + c, we can see that the slope (m) is 1.
Now we know the slope of both tangents, we can use the fact that the tangents to a parabola at a point are perpendicular to the line joining the focus and that point.
So, we need to find the equations of two lines with known slopes (from the given tangents) that are perpendicular to the line joining the focus of the parabola (1, 2).
The line joining the focus (1, 2) to a point (x, y) on the parabola has the slope: (y - 2) / (x - 1).
Since the tangents are perpendicular to this line, their slopes will be negative reciprocals of this slope.
For the first tangent: slope * (-1) = -1 * (y - 2) / (x - 1) = -1 / (1 + (y-2) / (x-1))
For the second tangent: slope * (-1) = -1 * (y - 2) / (x - 1) = -1 / (1 + (y-2) / (x-1))
Now, we have two equations in terms of x and y, and we can solve them to find the points of tangency.
By solving the system of equations, we get:
For the first tangent: y = 2x - 2
For the second tangent: y = -2x + 2
Now, let's find the point of tangency on each of the tangents by substituting these equations into the equation of the parabola.
The general equation of a parabola with the focus (h, k) is given by the following equation:
(x - h)^2 = 4a(y - k)
In this case, the focus is at (1, 2). So substituting the point of tangency (x, y) into the equation, we get:
(x - 1)^2 = 4a(y - 2)
Let's substitute the point (x, y) from each tangent equation into this equation.
For the first tangent (y = 2x - 2):
(x - 1)^2 = 4a(2x - 4)
For the second tangent (y = -2x + 2):
(x - 1)^2 = 4a(-2x + 4)
Now, let's simplify these equations and solve for a.
Expanding the squares:
(x^2 - 2x + 1) = 8ax - 16a
(x^2 - 2x + 1) = -8ax + 16a
Rearranging the equations:
x^2 - (8a + 2)x + (16a + 1) = 0
x^2 + (8a + 2)x - (16a + 1) = 0
Comparing this equation with the general equation of a quadratic ax^2 + bx + c = 0, we get:
a = 1
b = (8a + 2)
c = -(16a + 1)
Substituting a = 1 into b and c:
b = (8 + 2) = 10
c = -(16 + 1) = -17
Now, we can use the formula for the length of the latus rectum:
Length of latus rectum = 4a
Substituting a = 1, we get:
Length of latus rectum = 4 * 1 = 4
Therefore, the length of the latus rectum is 4.