The joint equation of bisector of angles between the lines given by $5 x^{2} + 4 x y - y^{2} = 0$ is

x^{2} + 3 x y + y^{2} = 0

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x^{2} - 3 x y + y^{2} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + 3 x y - y^{2} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} - 3 x y - y^{2} = 0

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Solution

The correct option is B $x^{2} - 3 x y - y^{2} = 0$
Consider $5 x^{2} + 4 x y - y^{2} = 0$

$\Rightarrow (5 x - y) (x + y) = 0$

$\Rightarrow 5 x - y = 0, x + y = 0$

Let $P (x, y)$ be any point on one edge bisector. Since the focuses on the point bisectors are equi - inaccessible from both the lines, the separation of $P (x, y)$ from the line $(5 x - y = 0) =$ the separation of $P (x, y)$ from the line $(x + y = 0)$
$\frac{5 x - y}{\sqrt{25 + 1}} = \frac{x + y}{\sqrt{1 + 1}}$

$\Rightarrow \frac{5 x - y}{\sqrt{26}} = \frac{x + y}{\sqrt{2}}$

$\Rightarrow \frac{(5 x - y)^{2}}{26} = \frac{(x + y)^{2}}{2}$

$\Rightarrow 2 (5 x - y)^{2} = 26 (x + y)^{2}$

$\Rightarrow 2 (25 x^{2} + y^{2} - 10 x y) = 26 (x^{2} + y^{2} + 2 x y)$

$\Rightarrow 50 x^{2} + 2 y^{2} - 20 x y = 26 x^{2} + 26 y^{2} + 52 x y$

$\Rightarrow 24 x^{2} - 24 y^{2} - 72 x y = 0$

$\Rightarrow x^{2} - y^{2} - 3 x y = 0$

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x^{2} - 3 x y + y^{2} = 0

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α + β - γ =

x^{2} + y^{2} - 3 = 3 x y

2 x^{2} - 6 + y^{2} = 0

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{(x - y)^{2} + 3 x y}] + [\frac{(x + y)^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x 3 - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

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