Question

From the point P(1,√3) on the circle x2+y2= a tangent is drawn to the hyperbola x24−y21=1 which meets its transverse axis at Q. From Q a line is drawn parallel to conjugate axis, which cuts the hyperbola at R above the x-axis, then PR equals.

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Solution

For the given hyperbola, we find that a=2, b=1Let the equation of the tangent be y=mx±√a2m2+b2 which is the standard form of the equation of the tangent to a hyperbola through a point.It passes through the point P=(1,√3). So, we have:√3=m±√4m2+1On solving, we get m=1−1√3=0.423The tangent cuts the x-axis at the point Q whose coordinates are given by:(−√a2m2+b2m,0)=(−3.1,0)The point R lies above the x-axis and on the left branch of the hyperbola, so:So, substituting x=−√a2m2+b2m into the equation of the hyperbola, and solving for the positive value of y, we get: y=b2am=1.183. Thus, the coordinates of the point R are: (−3.1,1.183)So, the length of the line PR is √(√3−1.183)2+(1+3.1)2=4.14

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