In the following figure drawn, $Δ X Y Z$ is a right triangle, right angled at $Y$ such that $Y Z = b$ and $A (Δ X Y Z) = a$ .
If $Y P ⊥ X Z$ , then show that $Y P = \frac{2 a \cdot b}{\sqrt{b^{4} + 4 a^{2}}}$

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Solution

$\frac{1}{2} \times Y Z \times X Y = a$
$\Rightarrow X Y = \frac{2 a}{Y Z}$
$\Rightarrow X Y = \frac{2 a}{b}$
Let $∠ P Y Z = 0$ then $∠ P Z Y = 90 - θ$
$= 90 - ∠ Y X Z$
$\Rightarrow ∠ Y X Z = θ$
$cos θ$ in $∠ Z Y P$ of $△ P Y Z$
$cos θ = \frac{P Y}{Y Z} = \frac{P Y}{b} - (1)$
$cos θ$ in $∠ P X Y$ of $△ X Y Z$
$cos θ = \frac{X Y}{X Z} = \frac{\frac{2 a}{b}}{\sqrt{b^{2} + {(\frac{2 a}{b})}^{2}}} - (2)$
Combining $(1)$ and $(2)$
$\frac{P Y}{b} = \frac{\frac{2 a}{b}}{\sqrt{\frac{b^{4} + 4 a^{2}}{b^{2}}}} = \frac{2 a}{\sqrt{b^{4} + 4 a^{2}}}$
$P Y = \frac{2 a b}{\sqrt{b^{4} + 4 a^{2}}}$

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