STATEMENT-1 : An equation of a common tangent to the parabola [y^2 = 16 3 x] and the ellip…

Question

STATEMENT-1 : An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$ and the ellipse $2{x^2} + {y^2} = 4$ is $$y = 2x + 2\sqrt 3 $$
STATEMENT-2 :If line $$y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)$$ is a common tangent to the parabola $${y^2} = 16\sqrt {3x} $$and the ellipse $2{x^2} + {y^2} = 4$, then $m$ satisfies ${m^4} + 2{m^2} = 24$

Accepted Answer

Given equation of ellipse is $2{x^2} + {y^2} = 4$

$$ \Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$$

Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$ is

$$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

(  as equation of tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

is $y=mx+c$ where $$c =  \pm \sqrt {{a^2}{m^2} + {b^2}} $$  )

Now, Equation of tangent to the parabola

$${y^2} = 16\sqrt 3 x$$ is $$y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$$

(  as equation of tangent to the parabola

${y^2} = 4ax$ is $$y = mx + {a \over m}$$  )

On comparing $(1)$ and $(2),$ we get

$${{4\sqrt 3 } \over m} =  \pm \sqrt {2{m^2} + 4} $$

Squaring on both the sides, we get

$$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$$

$$ \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$$

$$ \Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$$

$$ \Rightarrow {m^2} = 4$$ ( as $${m^2} \ne  - 6$$  ) $$ \Rightarrow m =  \pm 2$$

$$ \Rightarrow $$ Equation of common tangents are $$y =  \pm 2x \pm 2\sqrt 3 $$

Thus, statement - $1$ is true.

Statement - $2$ is obviously true.

Solutions · 1