Answer
454.5k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)