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?
[answer:4rt2]

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$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{of} \mathrm{m} \mathrm{slope} \mathrm{to} \mathrm{cirlce} {\mathrm{x}}^2+{\mathrm{y}}^2=16 \mathrm{is} \\ \mathrm{y}=\mathrm{mx}+4\sqrt{1+{\mathrm{m}}^2} \\ \mathrm{It} \mathrm{passes} \mathrm{through} \mathrm{P}\left(0,\mathrm{h}\right) \\ \mathrm{h}=4\sqrt{1+{\mathrm{m}}^2} \\ \Rightarrow \sqrt{1+{\mathrm{m}}^2}=\frac{\mathrm{h}}{4} \\ \Rightarrow 1+{\mathrm{m}}^2=\frac{{\mathrm{h}}^2}{16} \\ \Rightarrow {\mathrm{m}}^2=\frac{{\mathrm{h}}^2}{16}-1 \\ \Rightarrow {\mathrm{m}}^2=\frac{\sqrt{{\mathrm{h}}^2-16}}{16} \\ \Rightarrow \mathrm{m}=\pm \frac{\sqrt{{\mathrm{h}}^2-16}}{4} \\ \mathrm{Hence} \mathrm{equation} \mathrm{tangents} \mathrm{are} \\ \mathrm{y}=\mathrm{mx}+4\sqrt{1+{\mathrm{m}}^2} \mathrm{gives} \\ \mathrm{y}=\frac{\sqrt{{\mathrm{h}}^2-16}}{4}\mathrm{x}+\frac{\mathrm{h}}{4} \mathrm{and} \mathrm{y}=-\frac{\sqrt{{\mathrm{h}}^2-16}}{4}\mathrm{x}+\frac{\mathrm{h}}{4} \\ \mathrm{For} \mathrm{x}-\mathrm{axis} \mathrm{y}=0 \\ 0=\frac{\sqrt{{\mathrm{h}}^2-16}}{4}\mathrm{x}+\frac{\mathrm{h}}{4} \mathrm{and} 0=-\frac{\sqrt{{\mathrm{h}}^2-16}}{4}\mathrm{x}+\frac{\mathrm{h}}{4} \\ \Rightarrow \mathrm{x}=-\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}} \mathrm{or} \mathrm{x}=\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}} \\ \mathrm{Hence} \mathrm{A}\left(-\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}},0\right) \mathrm{and} \mathrm{B}\left(\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}},0\right) \\ \mathrm{AB}=\mathrm{OA}+\mathrm{OB} \mathrm{where} \mathrm{O} \mathrm{is} \mathrm{origin} \\ =\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}}+\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}}=\frac{2\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}} \left[\mathrm{Take} \mathrm{h}>0 \mathrm{as} \mathrm{it} \mathrm{does} \mathrm{not} \mathrm{effect} \mathrm{result}\right] \\ \mathrm{Height} \mathrm{of} \mathrm{triangle} \mathrm{is} \mathrm{h} \left(\mathrm{As} \mathrm{P} \mathrm{is} \mathrm{P}\left(0,\mathrm{h}\right)\hspace{0.17em}\right) \\ \mathrm{Area}\left(\mathrm{A}\right)=\frac{1}{2}\times \frac{2\mathrm{h}}{\sqrt{{\mathrm{h}}^2-16}}\times \mathrm{h} \\ \mathrm{A}=\frac{{\mathrm{h}}^2}{\sqrt{{\mathrm{h}}^2-16}} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \frac{\mathrm{dA}}{\mathrm{dh}}=\frac{2\mathrm{h}.\sqrt{{\mathrm{h}}^2-16}-{\mathrm{h}}^2.{\displaystyle \frac{2\mathrm{h}}{2\sqrt{{\mathrm{h}}^2-16}}}}{\left({\mathrm{h}}^2-16\right)} \\ =\frac{2\mathrm{h}.\sqrt{{\mathrm{h}}^2-16}-{\displaystyle \frac{{\mathrm{h}}^3}{\sqrt{{\mathrm{h}}^2-16}}}}{\left({\mathrm{h}}^2-16\right)} \\ \mathrm{For} \mathrm{minima}/\mathrm{maxima} \frac{\mathrm{dA}}{\mathrm{dh}}=0 \\ \Rightarrow \frac{2\mathrm{h}.\sqrt{{\mathrm{h}}^2-16}-{\displaystyle \frac{{\mathrm{h}}^3}{\sqrt{{\mathrm{h}}^2-16}}}}{\left({\mathrm{h}}^2-16\right)}=0 \\ \Rightarrow 2\mathrm{h}.\sqrt{{\mathrm{h}}^2-16}-\frac{{\mathrm{h}}^3}{\sqrt{{\mathrm{h}}^2-16}}=0 \\ \Rightarrow 2\mathrm{h}.\sqrt{{\mathrm{h}}^2-16}=\frac{{\mathrm{h}}^3}{\sqrt{{\mathrm{h}}^2-16}} \\ \Rightarrow 2\left({\mathrm{h}}^2-16\right)={\mathrm{h}}^2 \\ \Rightarrow {\mathrm{h}}^2=32 \\ \Rightarrow \mathrm{h}=\pm 4\sqrt{2} \\ \mathrm{As} \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{area} \mathrm{will} \mathrm{tend} \mathrm{towards} \mathrm{infinite}, \mathrm{then} \mathrm{h}=\pm 4\sqrt{2} \\ \mathrm{must} \mathrm{give} \mathrm{minima}, \mathrm{hence} \\ \mathrm{h}=\pm 4\sqrt{2} \end{gathered}$$

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