The auxiliary equation of the lines passing t | vedclass

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

Question

(A) The equations of the required lines passing through the origin with slopes $m_1 = \sqrt{3}+1$ and $m_2 = \sqrt{3}-1$ are $y = m_1 x$ and $y = m_2

Vedclass · Accepted Answer

$m^{2}-2 \sqrt{3} m+2=0$

Vedclass · Answer

(A) The equations of the required lines passing through the origin with slopes $m_1 = \sqrt{3}+1$ and $m_2 = \sqrt{3}-1$ are $y = m_1 x$ and $y = m_2 x$.Since the auxiliary equation for a pair of lines $y = m_1 x$ and $y = m_2 x$ is $(m - m_1)(m - m_2) = 0$,we substitute the given slopes:$(m - (\sqrt{3}+1))(m - (\sqrt{3}-1)) = 0$$m^2 - m(\sqrt{3}-1) - m(\sqrt{3}+1) + (\sqrt{3}+1)(\sqrt{3}-1) = 0$$m^2 - m(\sqrt{3}-1 + \sqrt{3}+1) + ((\sqrt{3})^2 - 1^2) = 0$$m^2 - m(2\sqrt{3}) + (3 - 1) = 0$$m^2 - 2\sqrt{3}m + 2 = 0$

The auxiliary equation of the lines passing through the origin and having slopes $\sqrt{3}+1$ and $\sqrt{3}-1$ is

Similar Questions

The joint equation of the pair of lines passing through the origin with slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is

The equation of the pair of straight lines perpendicular to the pair $2 x^2+3 x y+2 y^2+10 x+5 y=0$ and passing through the origin is $..........$

If the slope of one line of the pair of lines $2x^2 + hxy + 6y^2 = 0$ is thrice the slope of the other line,then $h =$

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}| = $

If an equation $hxy + gx + fy + c = 0$ represents a pair of lines,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module