TS EAMCET 5 May 2018 Shift 1 Solved Paper

The equation of hyperbola is expressed as,
4x2−y2=4a2

x2

a2

−

y2

4a2

=1
since, the equation of hyperbola can be homogenous with thehelp of x‌cos‌ϕ+y‌sin‌ϕ=P then,

x2

a2

−

y2

4a2

=(

x‌cos‌ϕ+y‌sin‌ϕ

P

)2
[x2(

1

a2

−

cos2ϕ

P2

)+x2(

1

a2

−

cos2ϕ

P2

) −

2xy‌cos‌ϕ‌sin‌ϕ

P2

]=0
Therefore,
Coefficient of x2+ Coefficient of y2=0

1

a2

−

cos2ϕ

P2

−

1

4a2

−

sin2ϕ

P2

=0
P=

2

√3

a
since, P is also length of perpendicular from (0,0) to line.
Thus, the radius of circle P is

2a

√3