The value of lambda such that lambda x^2-10 x | vedclass

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

Question

(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing the given equation $\lambda x^2 - 10xy + 12y^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing the given equation $\lambda x^2 - 10xy + 12y^2 + 5x - 16y - 3 = 0$ with the general form,we get: $a = \lambda, h = -5, b = 12, g = \frac{5}{2}, f = -8, c = -3$. The condition for the equation to represent a pair of straight lines is $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Substituting the values: $\lambda(12)(-3) + 2(-8)(\frac{5}{2})(-5) - \lambda(-8)^2 - 12(\frac{5}{2})^2 - (-3)(-5)^2 = 0$. $-36\lambda + 200 - 64\lambda - 75 + 75 = 0$. $-100\lambda + 200 = 0$. $100\lambda = 200$. $\lambda = 2$.

The value of $\lambda$ such that $\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0$ represents a pair of straight lines is:

Similar Questions

The three lines given by the combined equation $y^3 - 4x^2y = 0$ represent:

The variance of $20$ observations is $5$. If each observation is multiplied by $2$,then the new variance of the resulting observations is:

The set of all solutions of the inequation $x^2 - 2x + 5 \leq 0$ in $R$ is

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

$A$ particle is released from height $S$ from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of the Earth and the speed of the particle at that instant are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module