Two conjugate diameters of the ellipse x^2/a^2+y^2/b^2=1 are drawn to meet the directrix at P and Q. PM is drawn perpendicular to OQ and ON perpendicular to OP. Prove that PM and ON meet at a fixed point and state its coordinates.
To prove that PM and ON meet at a fixed point, we need to find the equation of the lines PM and ON, and then find the point of intersection.
Let's start by finding the equation of the line PM. We know that PM is perpendicular to OQ, and OQ is a chord of the ellipse. The equation of a line perpendicular to another line can be found using the slope-intercept form.
Step 1: Find the slope of the line OQ.
The line OQ passes through the origin (0,0) and point Q. Let's assume the coordinates of point Q are (x1, y1). The slope of the line passing through two points (x1, y1) and (0, 0) can be calculated as follows:
slope(OQ) = (y1 - 0)/(x1 - 0) = y1/x1
Step 2: Find the slope of the line PM.
Since PM is perpendicular to OQ, the slope of PM is the negative reciprocal of the slope of OQ. Therefore:
slope(PM) = -1/slope(OQ) = -x1/y1
Step 3: Find the equation of the line PM.
Using the point-slope form, where the slope is -x1/y1 and the point M is (a*cosθ, b*sinθ), the equation of the line PM can be written as:
y - b*sinθ = (-x1/y1) * (x - a*cosθ)
Simplifying this equation, we get:
y = (-x1/y1)x + (a*x1*cosθ)/y1 + b*sinθ
Now, let's find the equation of the line ON.
Step 4: Find the slope of the line ON.
ON is perpendicular to OP, and since OP is vertical (parallel to the y-axis), the slope of ON is 0.
Step 5: Find the equation of the line ON.
Since the slope of ON is 0, the equation of the line ON can be written as:
y = b*sinθ
Now that we have the equations of lines PM and ON, we can find their point of intersection.
Step 6: Solve the system of equations.
To find the point of intersection, we need to solve the system of equations:
y = (-x1/y1)x + (a*x1*cosθ)/y1 + b*sinθ
y = b*sinθ
Substituting y = b*sinθ in the first equation, we get:
b*sinθ = (-x1/y1)x + (a*x1*cosθ)/y1 + b*sinθ
Canceling out the common term, we have:
0 = (-x1/y1)x + (a*x1*cosθ)/y1
Rearranging the equation, we get:
(x1/y1)x = (a*x1*cosθ)/y1
Simplifying this equation, we find:
x = a*cosθ
Therefore, the point of intersection of lines PM and ON is (x, y) = (a*cosθ, b*sinθ), which is a point on the ellipse.
Since the point of intersection is determined by θ, and θ can vary for different positions of the two conjugate diameters, the coordinates of the fixed point of intersection are determined by the values of a and b.
Hence, the fixed point of intersection is given by (x, y) = (a*cosθ, b*sinθ), where θ is the angle between the two conjugate diameters.