Answer

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Hint: Consider a tangent expression \[y=mx+\dfrac{a}{m}\] for the given parabola. For being

a common tangent, the tangent expression we created for parabola should also act as

tangent to the circle. Thus, equating the distance between the center of the circle and the

tangent expression will lead us to find the common tangent.

The given circle \[2{{x}^{2}}+2{{y}^{2}}=5\]can be rewritten as:

\[{{x}^{2}}+{{y}^{2}}=\dfrac{5}{2};\]

Therefore, the center of this circle is \[\left( 0,0 \right)\] and the radius is

\[\sqrt{\dfrac{5}{2}}\]units.

For any parabola \[{{y}^{2}}=4ax;\] the tangent of the parabola with slope ‘m’ is considered

as \[y=mx+\dfrac{a}{m}\].

Subsequently, the tangent to the parabola \[{{y}^{2}}=4\sqrt{5}x\] can be taken as:

\[y=mx+\dfrac{\sqrt{5}}{m};\]

Which can be rewritten as:

\[my={{m}^{2}}x+\sqrt{5}\];

As it was mentioned that the parabola and circle have a common tangent:

Then, the distance between \[\left( 0,0 \right)\] and \[my={{m}^{2}}x+\sqrt{5}\] is equal to

the radius of the circle.

Therefore, applying the distance formula,

\[\dfrac{\left| {{m}^{2}}\left( 0 \right)-m\left( 0 \right)+\sqrt{5} \right|}{\sqrt{{{\left(

{{m}^{2}} \right)}^{2}}+{{\left( -m \right)}^{2}}}}=\dfrac{\sqrt{5}}{2}\]

\[\dfrac{\sqrt{5}}{\sqrt{{{m}^{4}}+{{m}^{2}}}}=\dfrac{\sqrt{5}}{2}\]

Now, let us do squaring on both sides to simplify further;

\[2=\left( {{m}^{4}}+{{m}^{2}} \right)\]

\[\Rightarrow {{m}^{4}}+{{m}^{2}}+2=0\]

\[({{m}^{2}}+2)({{m}^{2}}-1)=0\]

\[\therefore {{m}^{2}}=1,{{m}^{2}}=-2\]

As \[{{m}^{2}}=-2\] is not valid, we shall consider \[m=\pm 1\].

So, the common tangents can be:

\[y=x+\sqrt{5}\], by substituting \[m=1\].

\[y=-x-\sqrt{5}\], by substituting \[m=-1\].

Now, for statement II:

On Solving;

\[\begin{align}

& {{m}^{4}}-3{{m}^{2}}+2=0 \\

& {{m}^{4}}-2{{m}^{2}}-{{m}^{2}}+2=0 \\

& \left( {{m}^{2}}-2 \right)\left( {{m}^{2}}-1 \right)=0 \\

\end{align}\]

We do have \[{{m}^{2}}=1\] from this expression, but it is not the same expression as

statement I.

So, both the statements are correct, but statement II is not the correct explanation of

statement I.

Hence, option b is the correct answer.

Note: You may also start the procedure by considering a common tangent expression with

respect to the given circle.

Also, as a square of a number can never be negative, we ignored \[{{m}^{2}}=-2\] as a

solution.

a common tangent, the tangent expression we created for parabola should also act as

tangent to the circle. Thus, equating the distance between the center of the circle and the

tangent expression will lead us to find the common tangent.

The given circle \[2{{x}^{2}}+2{{y}^{2}}=5\]can be rewritten as:

\[{{x}^{2}}+{{y}^{2}}=\dfrac{5}{2};\]

Therefore, the center of this circle is \[\left( 0,0 \right)\] and the radius is

\[\sqrt{\dfrac{5}{2}}\]units.

For any parabola \[{{y}^{2}}=4ax;\] the tangent of the parabola with slope ‘m’ is considered

as \[y=mx+\dfrac{a}{m}\].

Subsequently, the tangent to the parabola \[{{y}^{2}}=4\sqrt{5}x\] can be taken as:

\[y=mx+\dfrac{\sqrt{5}}{m};\]

Which can be rewritten as:

\[my={{m}^{2}}x+\sqrt{5}\];

As it was mentioned that the parabola and circle have a common tangent:

Then, the distance between \[\left( 0,0 \right)\] and \[my={{m}^{2}}x+\sqrt{5}\] is equal to

the radius of the circle.

Therefore, applying the distance formula,

\[\dfrac{\left| {{m}^{2}}\left( 0 \right)-m\left( 0 \right)+\sqrt{5} \right|}{\sqrt{{{\left(

{{m}^{2}} \right)}^{2}}+{{\left( -m \right)}^{2}}}}=\dfrac{\sqrt{5}}{2}\]

\[\dfrac{\sqrt{5}}{\sqrt{{{m}^{4}}+{{m}^{2}}}}=\dfrac{\sqrt{5}}{2}\]

Now, let us do squaring on both sides to simplify further;

\[2=\left( {{m}^{4}}+{{m}^{2}} \right)\]

\[\Rightarrow {{m}^{4}}+{{m}^{2}}+2=0\]

\[({{m}^{2}}+2)({{m}^{2}}-1)=0\]

\[\therefore {{m}^{2}}=1,{{m}^{2}}=-2\]

As \[{{m}^{2}}=-2\] is not valid, we shall consider \[m=\pm 1\].

So, the common tangents can be:

\[y=x+\sqrt{5}\], by substituting \[m=1\].

\[y=-x-\sqrt{5}\], by substituting \[m=-1\].

Now, for statement II:

On Solving;

\[\begin{align}

& {{m}^{4}}-3{{m}^{2}}+2=0 \\

& {{m}^{4}}-2{{m}^{2}}-{{m}^{2}}+2=0 \\

& \left( {{m}^{2}}-2 \right)\left( {{m}^{2}}-1 \right)=0 \\

\end{align}\]

We do have \[{{m}^{2}}=1\] from this expression, but it is not the same expression as

statement I.

So, both the statements are correct, but statement II is not the correct explanation of

statement I.

Hence, option b is the correct answer.

Note: You may also start the procedure by considering a common tangent expression with

respect to the given circle.

Also, as a square of a number can never be negative, we ignored \[{{m}^{2}}=-2\] as a

solution.

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