NIMCET 2016 Question Paper with solutions

Question : 96 of 120

Marks: +1, -0

Two common tangents to the circle

x^2+y^2

2a^2

and parabola

y^2

= 8ax are

Solution:

Equation of tangent to parabola

y^2=8ax

y=mx+{2a}/m

This tangent is also touches the circle

x^2+y^2=2a^2

∴ Length of perpendicular = Radius

⇒{{2a}/m}/√{m^2+1}=√2a

⇒{4a^2}/{m^2(m^2+1)}=2a^2

⇒m^2(m^2+1)=2

⇒m^4+m^2-2=0

⇒(m^2-1)(m^2+2)=0

⇒m^2=1

∴m=±1

Hence, equation ot common tangent is

y=±(x+2a)

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