If m_{1} and m_{2} are slopes of the lines re | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation is of the form $Ax^{2} + 2Hxy + By^{2} = 0$,where $A = \sec^{2} 	heta - \sin^{2} 	heta$,$2H = -2 	an 	heta$,and $B = \sin^{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given equation is of the form $Ax^{2} + 2Hxy + By^{2} = 0$,where $A = \sec^{2} 	heta - \sin^{2} 	heta$,$2H = -2 	an 	heta$,and $B = \sin^{2} 	heta$. For a pair of lines $Ax^{2} + 2Hxy + By^{2} = 0$,the sum of slopes is $m_{1} + m_{2} = -\frac{2H}{B}$ and the product of slopes is $m_{1}m_{2} = \frac{A}{B}$. Here,$m_{1} + m_{2} = \frac{2 	an 	heta}{\sin^{2} 	heta}$ and $m_{1}m_{2} = \frac{\sec^{2} 	heta - \sin^{2} 	heta}{\sin^{2} 	heta}$. We know that $|m_{1} - m_{2}| = \sqrt{(m_{1} + m_{2})^{2} - 4m_{1}m_{2}}$. Substituting the values: $|m_{1} - m_{2}| = \sqrt{\left(\frac{2 	an 	heta}{\sin^{2} 	heta}ight)^{2} - 4\left(\frac{\sec^{2} 	heta - \sin^{2} 	heta}{\sin^{2} 	heta}ight)}$ $|m_{1} - m_{2}| = \sqrt{\frac{4 	an^{2} 	heta - 4(\sec^{2} 	heta - \sin^{2} 	heta) \sin^{2} 	heta}{\sin^{4} 	heta}}$ Using $	an^{2} 	heta = \frac{\sin^{2} 	heta}{\cos^{2} 	heta}$ and $\sec^{2} 	heta = \frac{1}{\cos^{2} 	heta}$: $|m_{1} - m_{2}| = \sqrt{\frac{4 \frac{\sin^{2} 	heta}{\cos^{2} 	heta} - 4(\frac{1}{\cos^{2} 	heta} - \sin^{2} 	heta) \sin^{2} 	heta}{\sin^{4} 	heta}} = \sqrt{\frac{4 \frac{\sin^{2} 	heta}{\cos^{2} 	heta} - 4 \frac{\sin^{2} 	heta}{\cos^{2} 	heta} + 4 \sin^{4} 	heta}{\sin^{4} 	heta}} = \sqrt{\frac{4 \sin^{4} 	heta}{\sin^{4} 	heta}} = \sqrt{4} = 2$.

If $m_{1}$ and $m_{2}$ are slopes of the lines represented by $(\sec^{2} \theta - \sin^{2} \theta) x^{2} - 2 \tan \theta xy + \sin^{2} \theta y^{2} = 0$,then $|m_{1} - m_{2}| = $

Similar Questions

If the lines represented by $ax^2 - bxy - y^2 = 0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis,then $\tan(\alpha + \beta) = $

For what value of $m$ can the equation $y^2 + 2xy + 2x + my - 3 = 0$ be factored into two linear factors?

The equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $3x + 4y - 5 = 0$ is

If $4ab = 3h^2$,then the ratio of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

Let $PQR$ be a right-angled isosceles triangle,right-angled at $P(2, 1)$. If the equation of the line $QR$ is $2x + y = 3$,then the equation representing the pair of lines $PQ$ and $PR$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module