If the bisector of angles represented by $a x^{2} + 2 h x y + b y^{2} = 0$ and $a^{^{'}} x^{2} + 2 h^{^{'}} x y + b^{^{'}} y^{2} = 0$ are same, then

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Solution

The correct option is A

Since bisector are same, therefore $\frac{a - b}{h} = \frac{a^{^{'}} - b^{^{'}}}{h^{^{'}}}$

$\Rightarrow (a - b) h^{^{'}} = (a^{^{'}} - b^{^{'}}) h$ .

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If the bisector of angles represented by ax2+2hxy+by2=0 and a′x2+2h′xy+b′y2=0 are same, then

The correct option is A Since bisector are same, therefore a−bh=a′−b′h′ ⇒(a−b)h′=(a′−b′)h.

If the bisector of angles represented by $a x^{2} + 2 h x y + b y^{2} = 0$ and $a^{^{'}} x^{2} + 2 h^{^{'}} x y + b^{^{'}} y^{2} = 0$ are same, then

The correct option is A

Since bisector are same, therefore $\frac{a - b}{h} = \frac{a^{^{'}} - b^{^{'}}}{h^{^{'}}}$

$\Rightarrow (a - b) h^{^{'}} = (a^{^{'}} - b^{^{'}}) h$ .