Tangent of acute angle between pair of straight lines $a x^{2} + 2 h x y + b y^{2}$ is given by $t a n θ = ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a - b} ∣ ∣ ∣$

True

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False

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Solution

The correct option is B
False

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

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

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

We know, acute angle between two lines

$T a n θ = ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣$

We need the value of $m_{1} - m_{2}$ .This can be find out using identity

$(m_{1} - m_{2})^{2} = (m_{1} + m_{2})^{2} - 4 m_{1} m_{2}$

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

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

$(m_{1} - m_{2})^{2} = \frac{4 (h^{2} - a b)}{b^{2}}$

$m_{1} - m_{2} = \frac{2 \sqrt{h^{2} - a b}}{b}$

So, tangent of acute angle

$t a n θ = \frac{\frac{2 \sqrt{h 62 - a b}}{b}}{a + \frac{a}{b}} = \frac{2 \sqrt{h^{2} - a b}}{a + b}$

$t a n θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$

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

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Similar questions

Tangent of acute angle between pair of straight lines $a x^{2} + 2 h x y + b y^{2}$ is given by $t a n θ = ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a - b} ∣ ∣ ∣$

The angle between the lines represented by the equation $a x^{2} + 2 h x y + b y^{2} = 0$ is given by

a x^{2} + 2 h x y + b y^{2} = 0

represents a pair of straight lines through origin & angle between them is given by

tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}

. If the lines are perpendicular then

a + b = 0

and the equation of bisectors is given by

\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h}

The general equation of second degree given by

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represent a pair of straight lines if

△ = 0

∣ ∣ ∣ ∣ \begin{matrix} a & h & g h & b & f g & f & c \end{matrix} ∣ ∣ ∣ ∣ = 0

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0

On the basis of above information answer the following question

The Joint equation of the bisectors of the angle between the lines represented by

a x^{2} + 2 h x y + b y^{2} = 0

Q. If

θ

is the acute angle between the lines represented by equation

a x^{2} + 2 h x y + b y^{2} = 0

, then prove that

tan θ = ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a + b} ∣ ∣ ∣, a + b \neq 0

Q. If

θ

is the measure of the acute angle between the lines represented by the equation

a x^{2} + 2 h x y + b y^{2} = 0

, then prove that

tan θ = ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a + b} ∣ ∣ ∣

where

a + b \neq 0

and

b \neq 0

.
Find the condition for coincident lines

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Tangent of acute angle between pair of straight lines ax2+2hxy+by2 is given by tanθ=∣∣∣2√h2−aba−b∣∣∣

Tangent of acute angle between pair of straight lines $a x^{2} + 2 h x y + b y^{2}$ is given by $t a n θ = ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a - b} ∣ ∣ ∣$