Question

$a{x}^2+2hxy+b{y}^2=0$ represents a pair of straight lines through origin & angle between them is given by

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$. If the lines are perpendicular then $a+b=0$ and the equation of bisectors is given by $\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

The general equation of second degree given by

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represent a pair of straight lines if $△=0$ or

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$ or $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

On the basis of above information answer the following question

If the angle between the lines represented by $6x^2+5xy-4y^2+7x+13y-3=0$ is ${\tan }^{-1}\left(m\right)$ & $a^2+b^2-ab-a-b+1\le 0$, then $5a+6b$ is equal to

• A. $\frac{1}{m}$
• B. $m$
• C. $\frac{m}{2}$
• D. $2m$

Answer 1

The correct option is D $2m$

Compare given equation with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ we have,

$a=6b=4h=\frac{5}{2}$

$\theta ={\tan }^{-1}\left(m\right)$ (given)

$={\tan }^{-1}\left\{\frac{2\sqrt{h^2-ab}}{a+b}\right\}={\tan }^{-1}\left\{\frac{2\sqrt{\frac{25}{4}+24}}{6-4}\right\}$

$={\tan }^{-1}\left\{\frac{11}{2}\right\}\therefore m=\frac{11}{2}$

$\Rightarrow (2m=11)$ .............(*)

Again $a^2+b^2-ab-a-b+1\le 0$

$\Rightarrow \frac{1}{2}(2a^2+2b^2-2ab-2a-2b+2)\le 0$

$\Rightarrow \frac{1}{2}\{{(a-b)}^2+{(a-1)}^2+{(b-1)}^2\}\le 0$

Which is possible only if $a-b=0,a-1=0,b-1=0i.e.a=b,a=1,b=1$

$\therefore 5a+6b=11=2m$ (Using *)

Hence choice (d) is the correct answer.