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Question

If $$\theta$$ is the acute angle between the lines represented by equation $$ax^2 + 2hxy + by^2 = 0$$, then prove that $$\tan \theta = \left | \dfrac{2 \sqrt{h^2 - ab}}{a + b}\right |, a + b \neq 0$$.


Solution

Let $$m_1 \  and \ m_2$$ be the slopes of the lines by the equation

Given, $$ax^2 + 2hxy + by^2 = 0$$            ....... (i)

Let $$y = m_1x$$ and $$y = m_2x$$

$$\therefore (y - m_1 x)(y - m_2x) = 0$$

and $$m_1m_2 x^2 - (m_1 + m_2) xy + y^2 = 0$$          ......... (ii)

Comparing (i) and (ii), we get

$$\dfrac{m_1 m_2}{a} = \dfrac{1}{b} = \dfrac{m_1 + m_2}{2h}$$

$$\therefore m_1 m_2 = \dfrac{a}{b}$$ and $$m_1 + m_2 = \dfrac{-2 h}{b}$$

Thus, $$(m_1 - m_2)^2 = (m_1 + m_2)^2  - 4m_1 m_2$$

$$(m_1 - m_2)^2 = \left( - \dfrac{2h}{b} \right )^2 - 4 \left(\dfrac{a}{b} \right )$$

$$(m_1-m_2)^2= \dfrac{4(h^2 - ab)}{b^2}$$

Let the angle between $$y = m_1 x$$ and $$y = m_2x $$ be $$\theta$$.

$$\tan  \theta = \left| \dfrac{m_1 - m_2}{1 + m_1m_2} \right |$$

$$= \left | \dfrac{\sqrt{m_1 - m_2}^2}{1 + m_1 m_2} \right |$$

$$= \left | \dfrac{\sqrt{\dfrac{4 (h^2 - ab)}{b^2}}}{1 + \dfrac{a}{b}} \right |$$

$$= \left | \dfrac{2 \sqrt{(h^2 - ab)}}{a + b} \right |$$, if $$a + b \neq 0$$

Mathematics

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