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Two lines (L_1 and L_2) are drawn from point $(\alpha,\alpha)$
making an angle 45º with the lines $L_{3} \equiv x + y - f(\alpha) = 0$
and $L_{4} \equiv x + y + f(\alpha) = 0$. L_1 intersects L_3 and L_4
at A and B and L_2 intersects L_3 and L_4 at C and D respectively
$(|2\alpha| > |f(\alpha)|)$. If the area of trapezium ABDC is
independent of$\alpha$. if $f(\alpha) = \lambda\alpha^{q}$, where
$\lambda$ is a constant, then $\ |q|\$ is
$L_{4} \equiv x + y + f(\alpha) = 0$ds lkFk 45º s.k cukrhgSA
$(|2\alpha| > |f(\alpha)|)$. leyEc prqHkZt ABDC =Qy $\alpha$ ls
Loar= gS rFkk $f(\alpha) = \lambda\alpha^{q}$ tgk¡ $\lambda$ vpj gks
rks $|q|\$dk eku gS&

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Distance between AC and vkSj BD nwjh
$AC = 2(OP - OE) = 2\left( \sqrt{2}\alpha - \frac{|f(\alpha)|}{\sqrt{2}} \right)$
and
So blfy, $AC + BD = 4\sqrt{2}\alpha$
Area of ABCD =Qy
$= \frac{1}{2}\sqrt{2}f(\alpha).4\sqrt{2}\alpha = 4\alpha f(\alpha)$
[IMAGE] If Area independent of $\alpha$, then $f(\alpha) = \frac{k}{\alpha}$

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx