Given : A circle, 2x^2 + 2y^2 = 5 and a parabola, [y^2 = 4 5 x].

Question

Given : A circle, $2{x^2} + 2{y^2} = 5$ and a parabola, $${y^2} = 4\sqrt 5 x$$.

Statement-1 : An equation of a common tangent to these curves is $$y = x + \sqrt 5 $$.
Statement-2 : If the line, $$y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)$$ is their common tangent, then $m$ satiesfies ${m^4} - 3{m^2} + 2 = 0$.

Accepted Answer

Let common tangent be

$$y = mx + {{\sqrt 5 } \over m}$$

Since, perpendicular distance from center of the circle to

the common tangent is equal to radius of the circle,

therefore $${{{{\sqrt 5 } \over m}} \over {\sqrt {1 + {m^2}} }} = \sqrt {{5 \over 2}} $$

On squaring both the side, we get

$${m^2}\left( {1 + {m^2}} \right) = 2$$

$$ \Rightarrow {m^4} + {m^2} - 2 = 0$$

$$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$

$$ \Rightarrow m =  \pm 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$  ( as $$m \ne  \pm \sqrt 2 $$ )

$$y =  \pm \left( {x + \sqrt 5 } \right),$$ both statements are correct as $$m =  \pm 1$$

satisfies the given equation of statement - $2.$

Solutions · 1