Answer
Verified
413.7k+ views
Hint:Let y = mx + c be the equation of common tangent to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$. Use the formula $c=\dfrac{a}{m}$ to get one equation. Then use the fact that for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle to find another equation. Solve these to get values of c and m. put those values in the equation of the line y = mx + c to get the final answer.
Complete step-by-step answer:
In this question, we need to find the equation(s) of the common tangent(s) to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$.
Let y = mx + c be the equation of common tangent to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$.
The condition for the line y = mx + c to be the tangent to the parabola ${{y}^{2}}=4ax$ is
$c=\dfrac{a}{m}$.
On comparing the parabola ${{y}^{2}}=2x$ with the parabola ${{y}^{2}}=4ax$, we get to know that
$a=\dfrac{1}{2}$.
Substituting this to the above expression, we will get the following:
$c=\dfrac{1}{2m}$ … (1)
Now, for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle.
From the equation of the circle, we will find that the radius of the circle given by ${{x}^{2}}+{{y}^{2}}+4x=0$ is 2 units and its centre is (-2, 0).
We know that the distance of a point (p, q) from a line ax + by +c = 0 is given by:
$\dfrac{|ap+bq+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Using this formula for the line y = mx + c and the point (-2, 0) and equating that to the radius of the circle, which is 2 units, we will get the following:
\[\dfrac{-2m+c}{\sqrt{1+{{m}^{2}}}}=2\] …(2)
Substituting equation (1) in equation (2), and then solving for m, we will get the following:
$m=\pm \dfrac{1}{2\sqrt{6}}$
Putting these in equation (1), we will get the following:
$c=\pm 2\sqrt{6}$
Substituting these values in the equation of the line y = mx + c .
So, the equations for the line y = mx + c are $x+2\sqrt{6}y+12=0$ and $x-2\sqrt{6}y+12=0$
So, the equations of the common tangents to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$ are $x+2\sqrt{6}y+12=0$ and $x-2\sqrt{6}y+12=0$.
Hence, options (b) and (c) are correct.
Note: In this question, it is very important to know the following formulae/ facts: The condition for the line y = mx + c to be the tangent to the parabola ${{y}^{2}}=4ax$ is $c=\dfrac{a}{m}$, for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle and that the distance of a point (p, q) from a line ax + by +c = 0 is given by: $\dfrac{|ap+bq+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.The equation $x^2+y^2+2gx+2fy+c=0$ represents a circle with centre (–g,–f) and radius ${\sqrt{g^2+f^2-c}}$. This is called the general equation of a circle.From given equation ${{x}^{2}}+{{y}^{2}}+4x=0$ comparing with general equation of circle $2g=4$ and $2f=0$ so centre will be $(-g,-f)$ i.e $(-2 ,0)$ and $radius$ =${\sqrt{2^2+0-0}}=2units$.
Complete step-by-step answer:
In this question, we need to find the equation(s) of the common tangent(s) to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$.
Let y = mx + c be the equation of common tangent to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$.
The condition for the line y = mx + c to be the tangent to the parabola ${{y}^{2}}=4ax$ is
$c=\dfrac{a}{m}$.
On comparing the parabola ${{y}^{2}}=2x$ with the parabola ${{y}^{2}}=4ax$, we get to know that
$a=\dfrac{1}{2}$.
Substituting this to the above expression, we will get the following:
$c=\dfrac{1}{2m}$ … (1)
Now, for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle.
From the equation of the circle, we will find that the radius of the circle given by ${{x}^{2}}+{{y}^{2}}+4x=0$ is 2 units and its centre is (-2, 0).
We know that the distance of a point (p, q) from a line ax + by +c = 0 is given by:
$\dfrac{|ap+bq+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Using this formula for the line y = mx + c and the point (-2, 0) and equating that to the radius of the circle, which is 2 units, we will get the following:
\[\dfrac{-2m+c}{\sqrt{1+{{m}^{2}}}}=2\] …(2)
Substituting equation (1) in equation (2), and then solving for m, we will get the following:
$m=\pm \dfrac{1}{2\sqrt{6}}$
Putting these in equation (1), we will get the following:
$c=\pm 2\sqrt{6}$
Substituting these values in the equation of the line y = mx + c .
So, the equations for the line y = mx + c are $x+2\sqrt{6}y+12=0$ and $x-2\sqrt{6}y+12=0$
So, the equations of the common tangents to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$ are $x+2\sqrt{6}y+12=0$ and $x-2\sqrt{6}y+12=0$.
Hence, options (b) and (c) are correct.
Note: In this question, it is very important to know the following formulae/ facts: The condition for the line y = mx + c to be the tangent to the parabola ${{y}^{2}}=4ax$ is $c=\dfrac{a}{m}$, for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle and that the distance of a point (p, q) from a line ax + by +c = 0 is given by: $\dfrac{|ap+bq+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.The equation $x^2+y^2+2gx+2fy+c=0$ represents a circle with centre (–g,–f) and radius ${\sqrt{g^2+f^2-c}}$. This is called the general equation of a circle.From given equation ${{x}^{2}}+{{y}^{2}}+4x=0$ comparing with general equation of circle $2g=4$ and $2f=0$ so centre will be $(-g,-f)$ i.e $(-2 ,0)$ and $radius$ =${\sqrt{2^2+0-0}}=2units$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE