A line through the origin meets the circle x2 + y2 = a2 at P and the h...
the equation of the tangents to the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q are
Where y = mx is intersecting line through (2, 0)
Let (h, k) be the point of intersection of these two lines
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A line through the origin meets the circle x2 + y2 = a2 at P and the h...
Y2 / b2 = 1 at Q. Prove that the rectangle contained by the segments OP and OQ is bisected by the x-axis.
Solution:
Let the equation of the line passing through the origin be y = mx, where m is the slope.
Then, the coordinates of point P on the circle are (a√(1+m2)/√(1+m2), am√(1/1+m2)), and the coordinates of point Q on the hyperbola are (b√(1+m2)/√(1-m2), bm√(1/1-m2)).
The distance between O and P is OP = a√(1+m2)/√(1+m2) = a, since the origin is the center of the circle.
The distance between O and Q is OQ = b√(1+m2)/√(1-m2).
The x-coordinate of point P is a√(1+m2)/√(1+m2) = a, and the x-coordinate of point Q is b√(1+m2)/√(1-m2).
Therefore, the length of the rectangle contained by segments OP and OQ is a * OQ = a * b√(1+m2)/√(1-m2).
To prove that this rectangle is bisected by the x-axis, we need to show that the y-coordinate of the midpoint of segment PQ is zero.
The midpoint of segment PQ has coordinates ((a√(1+m2)/√(1+m2) + b√(1+m2)/√(1-m2))/2, (am√(1/1+m2) + bm√(1/1-m2))/2).
Simplifying this expression, we get ((a+b)√(1+m2)/2√(1+m2)√(1-m2), (ab(m2-1))/(2m(1+m2)(1-m2))).
The y-coordinate of the midpoint is zero if and only if ab(m2-1) = 0.
Since a and b are both positive, this implies that m = ±1, which corresponds to the lines y = x and y = -x.
Therefore, the rectangle contained by segments OP and OQ is bisected by the x-axis when the line passing through the origin has slope ±1.
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