Question Statement Solve the given quadratic equation and find the acute angle between the lines represented by it:
x 2 + 2 x y sec β‘ x + y 2 = 0 x^2 + 2xy \sec x + y^2 = 0 x 2 + 2 x y sec x + y 2 = 0 Background and Explanation This is a quadratic equation in terms of x x x y y y y x \frac{y}{x} x y β sec β‘ = 1 cos β‘ \sec = \frac{1}{\cos} sec = c o s 1 β tan β‘ = sin β‘ cos β‘ \tan = \frac{\sin}{\cos} tan = c o s s i n β
Solution Step 1: Rewrite the equation
Divide through by x 2 x^2 x 2
y 2 x 2 + 2 x y sec β‘ x x 2 + x 2 x 2 = 0 \frac{y^2}{x^2} + \frac{2xy \sec x}{x^2} + \frac{x^2}{x^2} = 0 x 2 y 2 β + x 2 2 x y sec x β + x 2 x 2 β = 0 Simplify each term:
( y x ) 2 + 2 sec β‘ ( y x ) + 1 = 0 \left(\frac{y}{x}\right)^2 + 2 \sec \left(\frac{y}{x}\right) + 1 = 0 ( x y β ) 2 + 2 sec ( x y β ) + 1 = 0 Let y x = m \frac{y}{x} = m x y β = m
m 2 + 2 m sec β‘ + 1 = 0 m^2 + 2m \sec + 1 = 0 m 2 + 2 m sec + 1 = 0 Step 2: Solve the quadratic equation
Using the quadratic formula:
m = β 2 sec β‘ Β± ( 2 sec β‘ ) 2 β 4 ( 1 ) ( 1 ) 2 ( 1 ) m = \frac{-2\sec \pm \sqrt{(2\sec)^2 - 4(1)(1)}}{2(1)} m = 2 ( 1 ) β 2 sec Β± ( 2 sec ) 2 β 4 ( 1 ) ( 1 ) β β Substitute and simplify:
m = β 2 sec β‘ Β± 4 sec β‘ 2 β 4 2 m = \frac{-2\sec \pm \sqrt{4\sec^2 - 4}}{2} m = 2 β 2 sec Β± 4 sec 2 β 4 β β m = β 2 sec β‘ Β± 2 sec β‘ 2 β 1 2 m = \frac{-2\sec \pm 2\sqrt{\sec^2 - 1}}{2} m = 2 β 2 sec Β± 2 sec 2 β 1 β β Since sec β‘ 2 β 1 = tan β‘ 2 \sec^2 - 1 = \tan^2 sec 2 β 1 = tan 2
m = sec β‘ Β± tan β‘ m = \sec \pm \tan m = sec Β± tan Step 3: Interpret the slopes
Slopes of the lines are:
m 1 = sec β‘ + tan β‘ , m 2 = sec β‘ β tan β‘ m_1 = \sec + \tan, \quad m_2 = \sec - \tan m 1 β = sec + tan , m 2 β = sec β tan Step 4: Find the angle between the lines
The formula for the angle ΞΈ \theta ΞΈ m 1 m_1 m 1 β m 2 m_2 m 2 β
tan β‘ ΞΈ = β£ m 2 β m 1 1 + m 1 β
m 2 β£ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 \cdot m_2} \right| tan ΞΈ = β 1 + m 1 β β
m 2 β m 2 β β m 1 β β β Substitute m 1 = sec β‘ + tan β‘ m_1 = \sec + \tan m 1 β = sec + tan m 2 = sec β‘ β tan β‘ m_2 = \sec - \tan m 2 β = sec β tan
tan β‘ ΞΈ = β£ ( sec β‘ β tan β‘ ) β ( sec β‘ + tan β‘ ) 1 + ( sec β‘ + tan β‘ ) ( sec β‘ β tan β‘ ) β£ \tan \theta = \left| \frac{(\sec - \tan) - (\sec + \tan)}{1 + (\sec + \tan)(\sec - \tan)} \right| tan ΞΈ = β 1 + ( sec + tan ) ( sec β tan ) ( sec β tan ) β ( sec + tan ) β β Simplify the numerator:
tan β‘ ΞΈ = β£ β 2 tan β‘ 1 + ( sec β‘ 2 β tan β‘ 2 ) β£ \tan \theta = \left| \frac{-2\tan}{1 + (\sec^2 - \tan^2)} \right| tan ΞΈ = β 1 + ( sec 2 β tan 2 ) β 2 tan β β Using sec β‘ 2 β tan β‘ 2 = 1 \sec^2 - \tan^2 = 1 sec 2 β tan 2 = 1
tan β‘ ΞΈ = β£ β 2 tan β‘ 1 + 1 β£ \tan \theta = \left| \frac{-2\tan}{1 + 1} \right| tan ΞΈ = β 1 + 1 β 2 tan β β tan β‘ ΞΈ = β£ β tan β‘ 1 β£ = tan β‘ \tan \theta = \left| -\frac{\tan}{1} \right| = \tan tan ΞΈ = β β 1 tan β β = tan Thus, the acute angle ΞΈ \theta ΞΈ
ΞΈ = tan β‘ β 1 ( tan β‘ ) \theta = \tan^{-1}(\tan) ΞΈ = tan β 1 ( tan )
Quadratic Formula : m = β b Β± b 2 β 4 a c 2 a m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} m = 2 a β b Β± b 2 β 4 a c β β Relationship Between Slopes : tan β‘ ΞΈ = β£ m 2 β m 1 1 + m 1 β
m 2 β£ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 \cdot m_2} \right| tan ΞΈ = β 1 + m 1 β β
m 2 β m 2 β β m 1 β β β Trigonometric Identities :
sec β‘ = 1 cos β‘ \sec = \frac{1}{\cos} sec = c o s 1 β tan β‘ 2 = sec β‘ 2 β 1 \tan^2 = \sec^2 - 1 tan 2 = sec 2 β 1
Summary of Steps
Rewrite the equation in terms of y x \frac{y}{x} x y β
Solve the quadratic equation to find m 1 m_1 m 1 β m 2 m_2 m 2 β
Use the formula for the angle between two lines to calculate tan β‘ ΞΈ \tan \theta tan ΞΈ
Simplify using trigonometric identities to find the final acute angle.
