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 $a x^2 + 2 h x y

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 $a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0$,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 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: $2 \left( -36 - \frac{\lambda^2}{4} ight) + 5 \left( 15 - \frac{5 \lambda}{4} ight) + \frac{5}{2} \left( -\frac{5 \lambda}{2} - 30 ight) = 0$. $-72 - \frac{\lambda^2}{2} + 75 - \frac{25 \lambda}{4} - \frac{25 \lambda}{4} - 75 = 0$. $-\frac{\lambda^2}{2} - \frac{50 \lambda}{4} - 72 = 0$. $-\frac{\lambda^2}{2} - \frac{25 \lambda}{2} - 72 = 0$. Multiplying by $-2$: $\lambda^2 + 25 \lambda + 144 = 0$. $(\lambda + 9)(\lambda + 16) = 0$. So,$\lambda = -9$ or $\lambda = -16$. Given $|\lambda| < 16$,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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