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Question: The common tangent to the parabola \({y^2} = 4ax\) and \({x^2} = 4ay\) is \( {\text{A}}{\text...

The common tangent to the parabola y2=4ax{y^2} = 4ax and x2=4ay{x^2} = 4ay is
A. x+y+a=0 B. x+y−a=0 C. x−y+a=0 D. x−y−a=0  {\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 \\\

Explanation

Solution

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 y2=4ax{y^2} = 4ax is y=mx+amy = mx + \dfrac{a}{m},if it touches x2=4ay{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 x2=4a(mx+am){x^2} = 4a\left( {mx + \dfrac{a}{m}} \right) has equal roots or, mx2−4am2x−4a2=0m{x^2} - 4a{m^2}x - 4{a^2} = 0 has equal roots.
We know condition of equal roots D=0 i.e b2−4ac=0 ) D = 0 \\\ {\text{i}}{\text{.e }}{{\text{b}}^2} - 4ac = 0 \\\ )
⇒b2=16a2m4,4ac=−16a2m4\Rightarrow {b^2} = 16{a^2}{m^4},4ac = - 16{a^2}{m^4}
⇒16a2m4=−16a2m4⇒m=−1(∵m≠0)\Rightarrow 16{a^2}{m^4} = - 16{a^2}{m^4} \Rightarrow m = - 1\left( {\because m \ne 0} \right)
Putting m=−1m = - 1 in y=mx+amy = mx + \dfrac{a}{m}, we get y=−x−ay = - x - a
Or, x+y+a=0x + y + a = 0
Hence option A{\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.