If the pair of lines [ax^2 + 2( a + b )xy + by^2 = 0] lie along diameters of a circle and …

Question

If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then :

Accepted Answer

As per question area of one sector $=3$ area of another sector

$$ \Rightarrow $$ at center by one sector $$ = 3 \times $$ angle at center by another sector

Let one angle be $$\theta $$ then other $=30$

Clearly $$\theta  + 3\theta  = 180 \Rightarrow \theta  = {45^ \circ }$$

$$\therefore$$ Angle between the diameters represented by combined equation

$$a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0$$ is $${45^ \circ }$$

$$\therefore$$ Using $tan$ $$\theta $$ $$ = {{2\sqrt {{h^2} - ab} } \over {a + b}}$$

we get $$\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}$$

$$ \Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}$$

$$ \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)$$

$$ \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab$$

$$ \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0$$

Solutions · 1