Question

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 the line $y = mx + \dfrac{{4\sqrt 3 }}{m},\left( {m \ne 0} \right)$ is a common tangent to the parabola ${y^2} = 16\sqrt 3 x$ and the ellipse, $2{x^2} + {y^2} = 4$, then m satisfies ${m^4} + 2{m^2} = 24$
$\left( a \right)$ Statement 1 is false, statement 2 is true.
$\left( b \right)$ Statement 1 is true, statement 2 is true. Statement 2 is a correct explanation for statement 1.
$\left( c \right)$ Statement 1 is true, statement 2 is true. Statement 2 is not a correct explanation for statement 1.
$\left( d \right)$ Statement 1 is true, statement 2 is false.

Answer 1

Hint: In this particular question use the concept that the equation of tangent to the ellipse in slope form is given as, $y = mx \pm \sqrt {{{\left( {ma} \right)}^2} + {b^2}} $, where a and b are the length of semi major and semi minor axis respectively, when written in standard form, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Statement – 2: If the line $y = mx + \dfrac{{4\sqrt 3 }}{m},\left( {m \ne 0} \right)$ is a common tangent to the parabola ${y^2} = 16\sqrt 3 x$ and the ellipse, $2{x^2} + {y^2} = 4$, then m satisfies ${m^4} + 2{m^2} = 24$
As we all know that the standard equation of ellipse is,
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$, where a, and b are the length of semi major and semi minor axis respectively.
Now the given equation of ellipse is
$2{x^2} + {y^2} = 4$
Now convert it into standard form we have,
$ \Rightarrow \dfrac{{{x^2}}}{2} + \dfrac{{{y^2}}}{4} = 1$
So on comparing, ${a^2} = 2,{b^2} = 4$
Now as we know that the equation of tangent to the ellipse in slope form is given as, $y = mx \pm \sqrt {{{\left( {ma} \right)}^2} + {b^2}} $.
Now substitute the values of a and b we have,
$ \Rightarrow y = mx \pm \sqrt {2{m^2} + 4} $............... (1)
Now it is given that the line $y = mx + \dfrac{{4\sqrt 3 }}{m},\left( {m \ne 0} \right)$ is a common tangent to the parabola ${y^2} = 16\sqrt 3 x$ and the ellipse, $2{x^2} + {y^2} = 4$.
$ \Rightarrow y = mx + \dfrac{{4\sqrt 3 }}{m},\left( {m \ne 0} \right)$.................... (2)
So both the equations (1) and (2) are same so equate them we have,
$ \Rightarrow mx + \dfrac{{4\sqrt 3 }}{m} = mx \pm \sqrt {2{m^2} + 4} $
Now simplify we have,
$ \Rightarrow \dfrac{{4\sqrt 3 }}{m} = \pm \sqrt {2{m^2} + 4} $
Now squaring on both sides we have,
$ \Rightarrow {\left( {\dfrac{{4\sqrt 3 }}{m}} \right)^2} = {\left( { \pm \sqrt {2{m^2} + 4} } \right)^2}$
$ \Rightarrow \dfrac{{48}}{{{m^2}}} = 2{m^2} + 4$
$ \Rightarrow 48 = 2{m^4} + 4{m^2}$
Now divide by 2 throughout we have,
$ \Rightarrow {m^4} + 2{m^2} = 24$................ (3)
So this is the required answer which we have to show
Hence statement 2 is true.
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 $.
From equation (3), let ${m^2} = p$............... (4)
$ \Rightarrow {p^2} + 2p - 24 = 0$
Now factorize the above equation we have,
$ \Rightarrow {p^2} + 6p - 4p - 24 = 0$
$ \Rightarrow p\left( {p + 6} \right) - 4\left( {p + 6} \right) = 0$
$ \Rightarrow \left( {p + 6} \right)\left( {p - 4} \right) = 0$
$ \Rightarrow p = - 6,4$
Now from equation (4) we have,
$ \Rightarrow {m^2} = - 6,{m^2} = 4$
$ \Rightarrow {m^2} = - 6$ is not possible otherwise slope will be complex.
$ \Rightarrow {m^2} = 4$
$ \Rightarrow m = \pm 2$
Now from equation (2) we have,
$ \Rightarrow y = \pm \left( {2x + 2\sqrt 3 } \right)$
$ \Rightarrow y = \left( {2x + 2\sqrt 3 } \right),y = \left( { - 2x - 2\sqrt 3 } \right)$
Hence statement 1 is also true, and statement 2 is a correct explanation for statement 1.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the equation of tangent to the ellipse in slope form which is stated above, then equate this equation with the given tangent equation and simplify we will get the fourth order equation in terms of slope, then simplify this equation and substitute the value of slope in any tangent equation we will get the required answer.