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(A) \[a>0,b>0\]

(B) \[a>0,b<0\]

(C) \[a<0,b>0\]

(D) \[a<0,b<0\]

Answer

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Hint: Determine the tangent expression for the parabola at a variable point. As, both the curves touch each other externally, then equate the distance between the center of

circle to the determined tangent expression to the radius of the circle.

For the given parabola,\[{{y}^{2}}=4ax\] tangent at point \[P\left( a{{t}^{2}},2at \right)\] is given as,

\[\begin{align}

& y-2at=\dfrac{1}{t}\left( x-a{{t}^{2}} \right) \\

& x-yt+a{{t}^{2}}=0 \\

\end{align}\]

Now, consider the circle equation \[{{x}^{2}}+{{y}^{2}}+2bx=0\].

The given circle equation can be rewritten as:

\[{{x}^{2}}+{{y}^{2}}+2bx+{{b}^{2}}-{{b}^{2}}=0\].

\[{{\left( x+b \right)}^{2}}+{{y}^{2}}={{b}^{2}}\]

Circle center \[=\left( -b,0 \right)\] and radius \[=b\] units.

As the circle and parabola touch each other externally, we will have that distance between

\[x-yt+a{{t}^{2}}=0\] and \[\left( -b,0 \right)\] is equal to ‘b’, the radius of the circle.

Thus, applying the distance formula, we will obtain:

\[b=\left| \dfrac{-b+\left( -t \right)\left( 0 \right)+a{{t}^{2}}}{\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -t \right)}^{2}}}} \right|\]

Let us square on both the sides to simplify further:

\[{{b}^{2}}\left( 1+{{t}^{2}} \right)={{\left( a{{t}^{2}}-b \right)}^{2}}\]

Expanding the above square, we will have:

\[\begin{align}

& {{b}^{2}}+{{b}^{2}}{{t}^{2}}={{a}^{2}}{{t}^{4}}+{{b}^{2}}-2ab{{t}^{2}} \\

& 2ab+b={{a}^{2}}{{t}^{2}} \\

& {{t}^{2}}=\dfrac{b\left( 2a+b \right)}{{{a}^{2}}} \\

\end{align}\]

We know that \[{{t}^{2}}\ge 0\], as it is a perfect square.

\[\dfrac{b\left( 2a+b \right)}{{{a}^{2}}}\ge 0\]

\[b\left( 2a+b \right)\ge 0\]….. as \[{{a}^{2}}\ge 0\]

Case 1.

If \[b>0\]

Then \[2a+b>0\]

\[a>0,b>0\] is a possible solution.

Case 2.

If \[b<0\]

Then \[2a+b<0\]

\[a<0,b<0\] is also a possible solution.

So, it can be either \[a<0,b<0\] (or) \[a>0,b>0\]

Hence option A and D are the correct answers.

Option (A) and (D) are correct answers

Note: You can also consider the tangent equation of the circle first and follow the same

procedure. Also, remember that the square of any number is always positive.

circle to the determined tangent expression to the radius of the circle.

For the given parabola,\[{{y}^{2}}=4ax\] tangent at point \[P\left( a{{t}^{2}},2at \right)\] is given as,

\[\begin{align}

& y-2at=\dfrac{1}{t}\left( x-a{{t}^{2}} \right) \\

& x-yt+a{{t}^{2}}=0 \\

\end{align}\]

Now, consider the circle equation \[{{x}^{2}}+{{y}^{2}}+2bx=0\].

The given circle equation can be rewritten as:

\[{{x}^{2}}+{{y}^{2}}+2bx+{{b}^{2}}-{{b}^{2}}=0\].

\[{{\left( x+b \right)}^{2}}+{{y}^{2}}={{b}^{2}}\]

Circle center \[=\left( -b,0 \right)\] and radius \[=b\] units.

As the circle and parabola touch each other externally, we will have that distance between

\[x-yt+a{{t}^{2}}=0\] and \[\left( -b,0 \right)\] is equal to ‘b’, the radius of the circle.

Thus, applying the distance formula, we will obtain:

\[b=\left| \dfrac{-b+\left( -t \right)\left( 0 \right)+a{{t}^{2}}}{\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -t \right)}^{2}}}} \right|\]

Let us square on both the sides to simplify further:

\[{{b}^{2}}\left( 1+{{t}^{2}} \right)={{\left( a{{t}^{2}}-b \right)}^{2}}\]

Expanding the above square, we will have:

\[\begin{align}

& {{b}^{2}}+{{b}^{2}}{{t}^{2}}={{a}^{2}}{{t}^{4}}+{{b}^{2}}-2ab{{t}^{2}} \\

& 2ab+b={{a}^{2}}{{t}^{2}} \\

& {{t}^{2}}=\dfrac{b\left( 2a+b \right)}{{{a}^{2}}} \\

\end{align}\]

We know that \[{{t}^{2}}\ge 0\], as it is a perfect square.

\[\dfrac{b\left( 2a+b \right)}{{{a}^{2}}}\ge 0\]

\[b\left( 2a+b \right)\ge 0\]….. as \[{{a}^{2}}\ge 0\]

Case 1.

If \[b>0\]

Then \[2a+b>0\]

\[a>0,b>0\] is a possible solution.

Case 2.

If \[b<0\]

Then \[2a+b<0\]

\[a<0,b<0\] is also a possible solution.

So, it can be either \[a<0,b<0\] (or) \[a>0,b>0\]

Hence option A and D are the correct answers.

Option (A) and (D) are correct answers

Note: You can also consider the tangent equation of the circle first and follow the same

procedure. Also, remember that the square of any number is always positive.