Question-71410

Question Number 71410 by TawaTawa last updated on 15/Oct/19

Answered by MJS last updated on 15/Oct/19

A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} p \\ 0 \end{matrix}) C = (\begin{matrix} p \\ q \end{matrix}) D = (\begin{matrix} 0 \\ q \end{matrix})

D C : y = q

A C : y = \frac{q}{p} x

B E : y = - \frac{p}{q} x + \frac{p^{2}}{q}

E = B E \cap D C = (\begin{matrix} \frac{p^{2} - q^{2}}{p} \\ q \end{matrix})

F = A C \cap B E = (\begin{matrix} \frac{p^{3}}{p^{2} + q^{2}} \\ \frac{p^{2} q}{p^{2} + q^{2}} \end{matrix})

area (A B F) = \frac{p^{3} q}{2 (p^{2} + q^{2})}

area (B C F) = \frac{{p q}^{3}}{2 (p^{2} + q^{2})}

area (C E F) = \frac{q^{5}}{2 p (p^{2} + q^{2})}

area (A D E F) = \frac{q (p^{4} + p^{2} q^{2} - q^{4})}{2 p (p^{2} + q^{2})}

\frac{{p q}^{3}}{2 (p^{2} + q^{2})} = 5 \Rightarrow p^{2} + q^{2} = \frac{{p q}^{3}}{10}

\frac{q^{5}}{2 p (p^{2} + q^{2})} = 4 \Rightarrow p^{2} + q^{2} = \frac{q^{5}}{8 p}

\frac{{p q}^{3}}{10} = \frac{q^{5}}{8 p} \Rightarrow q = \frac{2 \sqrt{5}}{5} p \Rightarrow p = \frac{3}{2} \sqrt[4]{125} \land q = 3 \sqrt[4]{5}

\Rightarrow area (A B F) = \frac{p^{3} q}{2 (p^{2} + q^{2})} = \frac{25}{4} = 6.25

area (A D E F) = \frac{q (p^{4} + p^{2} q^{2} - q^{4})}{2 p (p^{2} + q^{2})} = \frac{29}{4} = 7.25

Commented by TawaTawa last updated on 15/Oct/19

I appreciate your time sir

Commented by TawaTawa last updated on 15/Oct/19

God bless you sir