Question

Match the entries of List - A and List - B.

Answer 1

(a) $2x^2-3xy-2y^2=0,\frac{\pi }{2}$
(b) $m+3m=-\frac{2h}{b^2},m.3m=\frac{a^2}{b^2}$
$\therefore 2m=-\frac{h}{b^2}⟹m=-\frac{h}{2b^2}$
$\therefore 3{(-\frac{h}{2b^2})}^2=\frac{a^2}{b^2}$ $\therefore h^2=\frac{4}{3}{a}^2{b}^2$
$h=\pm \frac{2}{\sqrt{3}}ab$.
(c) Here $a=12,b=2,c=\lambda ,f=-\frac{5}{2},g=\frac{11}{2},h=-5$
Putting these values in the condition
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$
we get $24\lambda +\frac{275}{2}-7.5-\frac{121}{2}-25\lambda =0$
$\therefore \lambda =2$
$12x^2-10xy+2y^2+11x-5y+2=0$
represents a pair of straight lines and proceeding as in part (a), their equations are
$4x-2y+1=0$ and $3x-y+2=0$.
The factors of $2^{nd}$ degree terms are $(2x-y)(6x-2y)$. Hence the two lines are $(2x-y+p)(6x-2y+q)$. Comparing the coefficient
$6p+2q=11,-2p-q=-5$ and $pq=\lambda$
The first two give $p=\frac{1}{2},q=4$ $\therefore pq=2=\lambda$
and the lines are $2x-y+\frac{1}{2}=0$
and $6x-2y+4=0$
The second degree terms cannot be factorized into two linear factors or else $h^2-ab=0-6=-ive$.
Hence the given equation does not represent a pair of lines whatever $\lambda$ may be.
(d) Clearly for $\lambda =2$ the second degree terms become $x^2-3xy+2y^2$ which can be factorized as $(x-2y)(x-y)$ and the two lines are $(x-2y+1)=0$ and $(x-y+2)=0$.