
The common tangent to the parabola ${y^2} = 4ax$ and ${x^2} = 4ay$ is
$
{\text{A}}{\text{. }}x + y + a = 0 \\
{\text{B}}{\text{. }}x + y - a = 0 \\
{\text{C}}{\text{. }}x - y + a = 0 \\
{\text{D}}{\text{. }}x - y - a = 0 \\
$
Answer
175.2k+ 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
JEE Main Physics Mock Test 2025

JEE Main Maths Mock Test 2025: FREE Online Mock Test Series

JEE Main Chemistry Mock Test 2025

JEE Main Hydrocarbons Mock Test 2025-26: Free Practice Online

JEE Main 2025-26 Mock Test: Organic Compounds Containing Nitrogen

JEE Main 2025-26 Mock Test: Organic Compounds Containing Halogens

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2025-26

NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2025-26

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2025-26

What is Hybridisation in Chemistry?

NCERT Solutions For Class 11 Maths Chapter 5 Linear Inequalities - 2025-26
