The value of lambda with |lambda|

Question

(B) The given equation is $2 x^2-10 x y+12 y^2+5 x+\lambda y-3=0$. Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx +

Vedclass · Accepted Answer

-$9$

Vedclass · Answer

(B) The given equation is $2 x^2-10 x y+12 y^2+5 x+\lambda y-3=0$. Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=2, h=-5, b=12, g=\frac{5}{2}, f=\frac{\lambda}{2}, c=-3$. For the equation to represent a pair of straight lines,the determinant of the matrix must be zero: $\begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$. Substituting the values: $\begin{vmatrix} 2 & -5 & 5/2 \ -5 & 12 & \lambda/2 \ 5/2 & \lambda/2 & -3 \end{vmatrix} = 0$. Expanding the determinant: $2(-36 - \frac{\lambda^2}{4}) + 5(15 - \frac{5\lambda}{4}) + \frac{5}{2}(\frac{-5\lambda}{2} - 30) = 0$. $-72 - \frac{\lambda^2}{2} + 75 - \frac{25\lambda}{4} - \frac{25\lambda}{4} - 75 = 0$. $-\frac{\lambda^2}{2} - \frac{50\lambda}{4} - 72 = 0$. Multiplying by $-2$: $\lambda^2 + 25\lambda + 144 = 0$. Factoring the quadratic: $(\lambda + 9)(\lambda + 16) = 0$. Thus,$\lambda = -9$ or $\lambda = -16$. Since the condition is $|\lambda| < 16$,the only valid solution is $\lambda = -9$.

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

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