Question

The angle between lines represented by the equations $a{x}^2+xy+b{y}^2=0$ will be ${35}^o$ if

• A. $a=1,b=6$
• B. $a=1,b=-6$
• C. $a=6,b=1$
• D. $Noneofthese$

Answer 1

The correct option is B $a=1,b=-6$

Given,

The equation of pair of strategies lines,

$a{x}^2+xy+b{y}^2=0$ - (i) and $\theta ={45}^{\circ }$

$\text{Equating equation (i) with standard}$

$\text{equation.}$

$a{x}^2+2hxy+b{y}^2=0$

$a,h=\frac{1}{2},b$

$\theta ={\tan }^{-1}\left(\frac{2\sqrt{h^2-ab}}{a+b}\right)$

$\tan {45}^{\circ }=\frac{2\sqrt{1/4-ab}}{a+b}$

$1(a+b)=\frac{2\sqrt{1-4ab}}{2}$

Squaring both sides

$a^2+b^2+2ab=1-4ab$

$a^2+b^2+6ab=1$ - (ii)

Putting option A in the above equation (ii)

$a=1,b=6,(1{)}^2+(6{)}^2+6\times 6=1$

$L\cdot H\cdot S\ne R\cdot H.S$

$B$ in equation(ii) $-a=1,b=-6$

$(1{)}^2+(-6{)}^2+6x-6=1$

L.H.S = R.H.S

:. Option B $=a=1,b=-6$