Mathematics | MHT CET 2020 | Discussion Page

If the equation ax²+2hxy+by²+2gx+2fy=0 has one line as the bisector of the angle between co-ordinate axes, then

$(a+b)^{2}=4(h^{2}+f^{2})$

$(a+b)^{2}=4(h^{2}+g^{2}+f^{2})$

$(a+b)^{2}=4h^{2}$

$(a+b)^{2}=4(h^{2}+g^{2})$

Option C

Given , equation of line

ax²+2hxy+by²+2gx+2fy=0

Let the slope of the line m₁ and m₂

$\therefore$ $m_{1}+m_{2}=-\frac{2h}{b}$

$m_{1}m_{2}=\frac{a}{b}$

Given, one line is bisector of angle between coordinate axes.

$\therefore$ $m_{1}=\tan\frac{\pi}{4}=1$

$\Rightarrow$ 1+m₂=-2h/b .........(i)

$m_{2}=\frac{a}{b}$ ......(ii)

From eqs. (i) and (ii) , we get

$1+\frac{a}{b}=\frac{-2h}{b}$

$\Rightarrow$ a+b=-2h

(a+b)² =4 h²

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