If $\theta$ is the acute angle between the pair of lines $\mathrm{H} \equiv a x^2-x y+b y^2=0, \tan \theta=5$ and $(1,-1)$ is a point on $\mathrm{H}=0$ then $a^2+a b+b^2=$

• A

  5
• B

  14
• Correct answer
  C

  7
• D

  13

Solution

having a point $(1,-1)$ satisfy $$ \Rightarrow \mathrm{a}+\mathrm{b}=-1 $$ We know that pair of straight lines passing through origin is $a x^2+2 h x y+b y^2=0$ On comparing (i) and (ii) $$ \begin{aligned} & \mathrm{h}=-\frac{1}{2} \\ & \because \tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \Rightarrow 5=\frac{2 \sqrt{\frac{1}{4}-\mathrm{ab}}}{-1} \Rightarrow \mathrm{ab}=-6 \\ & \text { Now } \mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2=(\mathrm{a}+\mathrm{b})^2-\mathrm{ab}=1+6=7 \end{aligned} $$

Hence, the correct answer is option: 'C'.