Question

If the line ax + by = 1 passes through point of intersection of $y=xtan\alpha +psec\alpha ,ysin({30}^{\circ }-\alpha )-xcos({30}^{\circ }-\alpha )=p$ and is inclined at ${30}^{\circ }$ with $y=xtan\alpha$, then the value of $a^2+b^2$ can be

• A. $\frac{1}{p^2}$
• B. $\frac{2}{p^2}$
  
• C. $\frac{3}{2p^2}$
  
• D. $\frac{3}{4p^2}$

Answer 1

The correct option is D

$\frac{3}{4p^2}$

Given, $ycos\alpha -xsin\alpha =p$

and $ysin({30}^{\circ }-\alpha )-xcos({30}^{\circ }-\alpha )=p$

are inclined at ${60}^{\circ }$ so line $ax+by=1$ can be acute angle bisector ...(i)

i.e., $ycos\alpha -xsin\alpha -p$

$=-\left(ysin\right({30}^{\circ }-\alpha )-xcos({30}^{\circ }-\alpha )-p)$

$\Rightarrow y[cos\alpha +sin({30}^{\circ }-\alpha \left)\right]-x[sin\alpha +cos({30}^{\circ }-\alpha \left)\right]=2p$ ...(ii)

From Eqs. (i) and (ii), we get

$\frac{b}{cos\alpha +sin({30}^{\circ }-\alpha )}=\frac{a}{(sin\alpha +cos({30}^{\circ }-\alpha \left)\right)}=\frac{1}{2p}$

$\Rightarrow \frac{\sqrt{a^2+b^2}}{\sqrt{2+1}}=\frac{1}{2p}$

$\Rightarrow a^2+b^2=\frac{3}{4p^2}$