Show that the latus rectum of the parabola $(a^{2} + b^{2}) (x^{2} + y^{2}) = (b x + a y - a b)^{2}$ is $2 a b \div \sqrt{a^{2} + b^{2}} .$

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Solution

Given equation of the parabola: $(a^{2} + b^{2}) (x^{2} + y^{2}) = (b x + a y - a b)^{2}$
Standard equation of parabola is of the form: $(d i s t a n c e f r o m A x i s)^{2} = \pm (L a t u s R e c t u m) \times (d i s t a n c e f r o m t a n g e n t a t V e r t e x) . . . e q n I$ , where Axis and tangent at Vertex are equations of some $m u t u a l l y p e r p e n d i c u l a r$ lines.
$(a^{2} + b^{2}) (x^{2} + y^{2}) = (b x + a y - a b)^{2}$
$\Rightarrow a^{2} x^{2} + a^{2} y^{2} + b^{2} x^{2} + b^{2} y^{2} = b^{2} x^{2} + a^{2} y^{2} + a^{2} b^{2} + 2 a b x y - 2 a^{2} b y - 2 a b^{2} x$
$\Rightarrow a^{2} x^{2} + b^{2} y^{2} - 2 a b x y = a^{2} b^{2} - 2 a^{2} b y - 2 a b^{2} x$
$\Rightarrow (a x - b y)^{2} = - a b (2 b x + 2 a y - a b)$
$\Rightarrow (a^{2} + b^{2}) \times {(\frac{a x - b y}{\sqrt{a^{2} + b^{2}}})}^{2} = - a b \times 2 (\sqrt{a^{2} + b^{2}}) \times ⎛ ⎜ ⎜ ⎝ \frac{2 b x + 2 a y - a b}{\sqrt{{(2 b)}^{2} + {(2 a)}^{2}}} ⎞ ⎟ ⎟ ⎠$
$\Rightarrow {(\frac{a x - b y}{\sqrt{a^{2} + b^{2}}})}^{2} = - \frac{2 a b}{(\sqrt{a^{2} + b^{2}})} \times ⎛ ⎜ ⎜ ⎝ \frac{2 b x + 2 a y - a b}{\sqrt{{(2 b)}^{2} + {(2 a)}^{2}}} ⎞ ⎟ ⎟ ⎠ . . . e q n I I$
Comparing $e q n I a n d e q n I I$ gives,
$L a t u s R e c t u m = \frac{2 a b}{\sqrt{a^{2} + b^{2}}}$

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