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Two common tangents to the circle $${x^2} + {y^2} = 2{a^2}$$ and parabola $${y^2} = 8ax$$ are :

JEE · Math · previous-year question

  1. A.$$x = \pm \left( {y + 2a} \right)$$
  2. B.$$y = \pm \left( {x + 2a} \right)$$correct
  3. C.$$x = \pm \left( {y + a} \right)$$
  4. D.$$y = \pm \left( {x + a} \right)$$

Answer

B. $$y = \pm \left( {x + 2a} \right)$$

Explanation

Any tangent to the parabola $${y^2} = 8ax$$ is $$y = mx + {{2a} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$ If $$(i)$$ is a tangent to the circle, $${x^2} + {y^2} = 2{a^2}$$ then, $$\sqrt {2a} = \pm {{2a} \over {m\sqrt {{m^2} + 1} }}$$ $$ \Rightarrow {m^2}\left( {1 + {m^2}} \right) = 2$$ $$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$ $$ \Rightarrow m = \pm 1.$$ So from $$(i),$$ $$y = \pm \left( {x + 2a} \right).$$

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