Answer
64.8k+ views
Hint: Whenever you get the questions of common tangent you have to write the equation of tangent to a given curve and then apply the condition of tangent for others. Here in case of parabola tangent means that equation has equal roots.so apply the condition and get a common tangent.
The equation of any tangent to ${y^2} = 4ax$ is $y = mx + \dfrac{a}{m}$,if it touches ${x^2} = 4ay$.
We know tangent touches at a single point so in case of parabola it is a quadratic equation and touches at single point means it has real and equal roots.
Then the equation ${x^2} = 4a\left( {mx + \dfrac{a}{m}} \right)$ has equal roots or, $m{x^2} - 4a{m^2}x - 4{a^2} = 0$ has equal roots.
We know condition of equal roots $
D = 0 \\
{\text{i}}{\text{.e }}{{\text{b}}^2} - 4ac = 0 \\ )$
$ \Rightarrow {b^2} = 16{a^2}{m^4},4ac = - 16{a^2}{m^4}$
$ \Rightarrow 16{a^2}{m^4} = - 16{a^2}{m^4} \Rightarrow m = - 1\left( {\because m \ne 0} \right)$
Putting $m = - 1$ in $y = mx + \dfrac{a}{m}$, we get $y = - x - a$
Or, $x + y + a = 0$
Hence option ${\text{A}}$ is correct.
Note: The key concept of solving questions of common tangent is first select a curve and write any general tangent to it and then apply the condition according to the second curve given in question. If the second curve is a circle then distance from center to tangent will be it’s radius but here in case of parabola equal roots will be conditioned.
The equation of any tangent to ${y^2} = 4ax$ is $y = mx + \dfrac{a}{m}$,if it touches ${x^2} = 4ay$.
We know tangent touches at a single point so in case of parabola it is a quadratic equation and touches at single point means it has real and equal roots.
Then the equation ${x^2} = 4a\left( {mx + \dfrac{a}{m}} \right)$ has equal roots or, $m{x^2} - 4a{m^2}x - 4{a^2} = 0$ has equal roots.
We know condition of equal roots $
D = 0 \\
{\text{i}}{\text{.e }}{{\text{b}}^2} - 4ac = 0 \\ )$
$ \Rightarrow {b^2} = 16{a^2}{m^4},4ac = - 16{a^2}{m^4}$
$ \Rightarrow 16{a^2}{m^4} = - 16{a^2}{m^4} \Rightarrow m = - 1\left( {\because m \ne 0} \right)$
Putting $m = - 1$ in $y = mx + \dfrac{a}{m}$, we get $y = - x - a$
Or, $x + y + a = 0$
Hence option ${\text{A}}$ is correct.
Note: The key concept of solving questions of common tangent is first select a curve and write any general tangent to it and then apply the condition according to the second curve given in question. If the second curve is a circle then distance from center to tangent will be it’s radius but here in case of parabola equal roots will be conditioned.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)