If the equation y^3 - 3x^2y + m(x^3 - 3xy^2) | vedclass

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

Question

(A) The given equation is $y^3 - 3xy^2m + 3x^2y(-1) - mx^3 = 0$. Dividing by $x^3$,we get $(y/x)^3 - 3m(y/x)^2 - 3(y/x) - m = 0$. Let $t = y/x = 	an

Vedclass · Accepted Answer

Lines are equally inclined to each other

Vedclass · Answer

(A) The given equation is $y^3 - 3xy^2m + 3x^2y(-1) - mx^3 = 0$. Dividing by $x^3$,we get $(y/x)^3 - 3m(y/x)^2 - 3(y/x) - m = 0$. Let $t = y/x = 	an 	heta$. Then $t^3 - 3mt^2 - 3t - m = 0$. This is a cubic equation in $t$. If we substitute $t = 	an 	heta$,the equation represents three lines $y = t_1x, y = t_2x, y = t_3x$. It can be shown that the angles between these lines are $\pi/3$ or $60^\circ$. Thus,the lines are equally inclined to each other.

If the equation $y^3 - 3x^2y + m(x^3 - 3xy^2) = 0$ represents three lines passing through the origin,then:

Similar Questions

The combined equation of a possible pair of adjacent sides of a square with area $16 \text{ square units}$ whose centre is the point of intersection of the lines $x+2y-3=0$ and $2x-y-1=0$ is

If the sum of the slopes of the lines represented by $x^2 - 2xy \tan \theta - y^2 = 0$ is $4$,then $\theta =$

For $\ell \in R$,the equation $(2 \ell-3) x^2+2 \ell xy-y^2=0$ represents a pair of distinct lines

The product of the perpendicular distances from the origin to the pair of straight lines $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ is

The equation $(x - 5)^2 + (x - 5)(y - 6) - 2(y - 6)^2 = 0$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module