If y=m1x+c1 and y=m2x+c2, \[{m_1} \ne {m_2}\] are two common tangents of circle x2+y2=2 and parabola y2 = x, then the value of 8∣m1m2∣ is equal to :
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Official solution
Let tangent to y2=x be
\[y = mx + {1 \over {4m}}\]
For it being tangent to circle.
\[\left| {{{{1 \over 4}m} \over {\sqrt {1 + {m^2}} }}} \right| = \sqrt 2 \]
\[ \Rightarrow 32{m^4} + 32{m^2} - 1 = 0\]
\[ \Rightarrow {m^2} = {{ - 32 \pm \sqrt {{{(32)}^2} + 4(32)} } \over {64}}\]
\[ \Rightarrow 8{m_1}{m_2} = - 4 + 3\sqrt 2 \]