Let the tangents drawn to the circle x^2+y^2=16 from the point P(0,h) meet the x-axis at points A and B. If the are of triangle APB is minimum, then h is equal to?

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Equation of tangent of m slope to cirlce x2+y2=16 isy=mx+41+m2It passes through P0,hh=41+m21+m2=h41+m2=h216m2=h216-1m2=h2-1616m=±h2-164Hence equation tangents arey=mx+41+m2 givesy=h2-164x+h4 and y=-h2-164x+h4For x-axis y=00=h2-164x+h4 and 0=-h2-164x+h4x=-hh2-16 or x=hh2-16Hence A-hh2-16,0 and Bhh2-16,0AB=OA+OB where O is origin=hh2-16+hh2-16=2hh2-16 Take h>0 as it does not effect resultHeight of triangle is h As P is P0,hAreaA=12×2hh2-16×hA=h2h2-16Differentiate with respect to xdAdh=2h.h2-16-h2.2h2h2-16h2-16=2h.h2-16-h3h2-16h2-16For minima/maxima dAdh=02h.h2-16-h3h2-16h2-16=02h.h2-16-h3h2-16=02h.h2-16=h3h2-162h2-16=h2h2=32h=±42As maximum value of area will tend towards infinite, then h=±42must give minima, henceh=±42

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