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A line through the origin meets the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q.
Then locus of the point of intersections of tangent to the circle at P with the tangent at Q to the
hyperbola is the curve (a4 + 4y4)x2 = aK then K is equal to .
    Correct answer is '6'. Can you explain this answer?
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    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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    A line through the origin meets the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q.Then locus of the point of intersections of tangent to the circle at P with the tangent at Q to thehyperbola is the curve (a4 + 4y4)x2 = aK then K is equal to .Correct answer is '6'. Can you explain this answer?
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    A line through the origin meets the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q.Then locus of the point of intersections of tangent to the circle at P with the tangent at Q to thehyperbola is the curve (a4 + 4y4)x2 = aK then K is equal to .Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A line through the origin meets the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q.Then locus of the point of intersections of tangent to the circle at P with the tangent at Q to thehyperbola is the curve (a4 + 4y4)x2 = aK then K is equal to .Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A line through the origin meets the circle x2 + y2 = a2 at P and the hyperbola x2 – y2 = a2 at Q.Then locus of the point of intersections of tangent to the circle at P with the tangent at Q to thehyperbola is the curve (a4 + 4y4)x2 = aK then K is equal to .Correct answer is '6'. Can you explain this answer?.
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