If $a x^{2} - 2 h x y - a y^{2} = 0$ represents a pair of straight lines forming with $2 x + 3 y = 6$ an isosceles right angled triangle at the origin then $12 a^{2} + 5 h^{2} + 5 a + 12 h + 60$ is equal to

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Solution

Let $y = m x$ be one of the lines through the origin.
Since the triangle is isosceles right angled, this line makes an angle of $\pm \frac{π}{4}$ with $2 x + 3 y = 6$ whose slope is $- \frac{2}{3}$ .
So $tan (\pm \frac{π}{4}) = \frac{m + (2 / 3)}{1 + m (- 2 / 3)} = \frac{3 m + 2}{3 - 2 m}$
$\Rightarrow {(3 - 2 m)}^{2} = {(3 m + 2)}^{2}$
$\Rightarrow [3 - 2 (y / x)^{2}] = {[3 (y / x) + 2]}^{2} \dots \dots [∵ m = y / x]$
$\Rightarrow {(3 x - 2 y)}^{2} = {(3 y + 2 x)}^{2}$
$\Rightarrow 5 x^{2} - 24 x y - 5 y^{2} = 0$
This is same as $a x^{2} - 2 h x y - a y^{2} = 0$
$\Rightarrow a = 5$ and $h = 12$
So, $12 a^{2} + 5 h^{2} + 5 a + 12 h + 60$
$= 12 \times 25 + 5 \times 144 + 25 + 144 + 60 = 300 + 720 + 229 = 1249$

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