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If $x^2+2hxy+y^2=0$ represents the equations of the stright lines through the origin which make an angle of $\alpha$ with the stright line $y+x=0,$ Then $\cot(\alpha)=$

What i try

$$\bigg(\frac{y}{x}\bigg)^2+2h\bigg(\frac{y}{x}\bigg)+1=0$$

Let $\displaystyle \frac{y}{x}=m.$ Then $m^2+2hm+1=0$ has a roots $m_{1},m_{2}$,

$$ m_{1}+m_{2}=-2h,\>\>\>\>\>m_{1}\cdot m_{2}=1$$

I am trying to use $\tan$gent of angle between two lines. But did not know what i can take i.e $m_{1},m_{2}$

How do i solve it Help me please

Quanto
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jacky
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1 Answers1

2

Note that, from the given angle $\alpha$ and the line $x+y=0$ (with its tangent angle $\frac{3\pi}4$),

$$m_1 = \tan(\frac{3\pi}4 - \alpha),\>\>\>\>\> m_2 = \tan(\frac{3\pi}4 + \alpha)$$

Then,

$$-2h=m_1+m_2 = \tan(\frac{3\pi}4 - \alpha)+ \tan(\frac{3\pi}4 + \alpha) =-\frac{2(1+\tan^2\alpha)}{1-\tan^2\alpha}$$

which leads to $\cot^2\alpha = \frac{h+1}{h-1}$.

Quanto
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