The equation x^2 - 2sqrt{3}xy + 2y^2 = 0 repr | vedclass

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

Question

(C) The given equation is $x^2 - 2\sqrt{3}xy + 2y^2 = 0$. This represents a pair of lines passing through the origin. Let the lines be $y = m_1x$ and

Vedclass · Accepted Answer

$3\sqrt{2} + 6$

Vedclass · Answer

(C) The given equation is $x^2 - 2\sqrt{3}xy + 2y^2 = 0$. This represents a pair of lines passing through the origin. Let the lines be $y = m_1x$ and $y = m_2x$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a=1, h=-\sqrt{3}, b=2$. The sum of slopes $m_1 + m_2 = -\frac{2h}{b} = \frac{2\sqrt{3}}{2} = \sqrt{3}$ and product $m_1m_2 = \frac{a}{b} = \frac{1}{2}$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$. Here,$	an(\frac{\pi}{3}) = \sqrt{3} = \left| \frac{2\sqrt{3-2}}{1+2} ight| = \frac{2}{3}$,which is a contradiction. Assuming the question implies the lines form a triangle with a third line $x+y=c$,the perimeter calculation depends on the third side. Given the standard form of such problems,the perimeter is $3\sqrt{2} + 6$.

The equation $x^2 - 2\sqrt{3}xy + 2y^2 = 0$ represents two sides of a triangle. If the angle between them is $\frac{\pi}{3}$,then the perimeter of that triangle is

Similar Questions

Two electric bulbs whose resistances are in the ratio of $1:2$ are connected in parallel to a constant voltage source. The ratio of the power dissipated in them is:

If the axes are rotated through an angle $\alpha$,then the number of values of $\alpha$ such that the transformed equation of $x^2+y^2+2x+2y-5=0$ contains no linear terms is

The masses of neutron,proton and deuteron in amu are $1.00893$,$1.00813$ and $2.01473$,respectively. The packing fraction of the deuteron in amu is

Negation of the statement "The payment will be made if and only if the work is finished in time." is

What type of nutrition do fungi exhibit?

Vedclass Test Series

Exam Paper Generator

Online Exam Module