Question

A variable line is drawn through origin ${}^{\prime }{O}^{\prime }$. Two points $A$ and $B$ are taken on the line in $1^{\text{st}}$ quadrant such that $OA=1$ unit and $OB=2$ units. Through points $A$ and $B$ two lines are drawn which making an equal angle $\alpha$ with the line $AB$. Then, the locus of point of intersection of these lines is

• A. $x^2+y^2=\frac{9+{\tan }^2\alpha }{4}$
• B. $x^2-y^2=\frac{9-{\tan }^2\alpha }{4}$
• C. $x^2+y^2=\frac{9+{\tan }^2\alpha }{2}$
• D. $x^2+y^2=\frac{9+2{\tan }^2\alpha }{4}$

Answer 1

The correct option is A $x^2+y^2=\frac{9+{\tan }^2\alpha }{4}$

Let $C$ be mid point of $AB$


Given that, $AB=1$
$\Rightarrow AC=CB=\frac{1}{2}\Rightarrow OC=\frac{3}{2}\text{and}PC=\frac{1}{2}\tan \alpha \Rightarrow O{P}^2=O{C}^2+P{C}^2\Rightarrow x^2+y^2={\left(\frac{3}{2}\right)}^2+\frac{1}{4}{\tan }^2\alpha$
$\therefore x^2+y^2=\frac{9+{\tan }^2\alpha }{4}$
is the required locus