Question

Tangent of acute angle between pair of straight lines $a{x}^2+2hxy+b{y}^2$ is given by $tan\theta =\left|\frac{2\sqrt{h^2-ab}}{a-b}\right|$

• A. True
• B. False

Answer 1

The correct option is B

False

Given equation $a{x}^2+2hxy+b{y}^2$ =0

Let's two equations passing through origin with slope $m_1$ And $m_2$ be

Y=$m_1$ x and y=$m_2$x

Pair of straight line

($m_1$x-y)($m_2$x-y)=0

$m_1$$m_2{x}^2$ - ($m_1$ + $m_2$)xy + $y^2$=0

Comparing this equation with given equation

$\frac{a{x}^2}{b}+\frac{2h}{b}xy+y^2=0$

$m_1{m}_2=\frac{a}{b}$ $m_1+m_2=\frac{-2h}{b}$

We know, acute angle between two lines

$Tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

We need the value of $m_1-m_2$.This can be find out using identity

$(m_1-m_2{)}^2=(m_1+m_2{)}^2-4m_1{m}_2$

= $(\frac{-2h}{b}{)}^2-\frac{4a}{b}$

= $\frac{4h^2}{b^2}-\frac{4a}{b}$

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

$m_1-m_2=\frac{2\sqrt{h^2-ab}}{b}$

So, tangent of acute angle

$tan\theta =\frac{\frac{2\sqrt{h62-ab}}{b}}{a+\frac{a}{b}}=\frac{2\sqrt{h^2-ab}}{a+b}$

$tan\theta =\frac{2\sqrt{h^2-ab}}{a+b}$

So, given statement is false because in denominator its given a - b.