A and B are fixed points with coordinates (0, | vedclass

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(A) Let $	heta = \angle APB$. The coordinates are $A(0, a)$, $B(0, b)$, and $P(x, 0)$.Using the law of cosines in $	riangle APB$:$AB^2 = PA^2 + PB^2

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$x^2 = ab$

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(A) Let $	heta = \angle APB$. The coordinates are $A(0, a)$, $B(0, b)$, and $P(x, 0)$.Using the law of cosines in $	riangle APB$:$AB^2 = PA^2 + PB^2 - 2(PA)(PB) \cos 	heta$$(a - b)^2 = (x^2 + a^2) + (x^2 + b^2) - 2 \sqrt{x^2 + a^2} \sqrt{x^2 + b^2} \cos 	heta$$a^2 - 2ab + b^2 = 2x^2 + a^2 + b^2 - 2 \sqrt{x^2 + a^2} \sqrt{x^2 + b^2} \cos 	heta$$2 \sqrt{x^2 + a^2} \sqrt{x^2 + b^2} \cos 	heta = 2x^2 + 2ab$$\cos 	heta = \frac{x^2 + ab}{\sqrt{x^2 + a^2} \sqrt{x^2 + b^2}}$To maximize $	heta$, we minimize $\cos 	heta$. Let $f(x) = \cos^2 	heta = \frac{(x^2 + ab)^2}{(x^2 + a^2)(x^2 + b^2)}$.Setting the derivative to zero or using the property that for a fixed base $AB$ on the $y$-axis, the angle subtended at $P(x, 0)$ is maximum when the circle passing through $A$ and $B$ is tangent to the $x$-axis at $P$, we get $x^2 = ab$.

$A$ and $B$ are fixed points with coordinates $(0, a)$ and $(0, b)$ where $a > 0$ and $b > 0$. $P$ is a variable point $(x, 0)$ on the $x$-axis. If the angle $\angle APB$ is maximum, then:

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