 The question says parabolas y square is equal to 4a x minus c1 and x square is equal to 4a y minus c2 Where c1 and c2 are variables Right now friends here c1 c2 are like parameters parameters means they are constants only but they keep on changing So as the c1 changes the parabola will change right so as the c2 change this parabola will also change Right, so as we all know the position of the vertex of the parabola will keep on changing Now these two parabola are variable parabolas because they are linked to the parameters c1 and c2 Now the question is asking that if these two parabolas touch each other What is the locus of their point of contact? What are the locus of their point of contact now friends? Let me tell you Jay main The the question based on locus are very very important Jay simply loves locus questions Okay, so how will I solve this question first thing in my mind that comes is First I will find the Derivative of both these parabolas, so I'll find dy by dx for parabola p1 okay, and I'll equate The dy by dx for both the parabolas Because if the two parabolas touch each other at a common point That means the slope of the tangent at that point of touching should be the same Yes or no, okay, so let's see what do we get and we'll assume the point of touching to be h comma k That's the point where the two parabolas touch each other and then we'll Generalize this then we'll generalize this point So with this roadmap in our mind, let us proceed towards solving the problem So let's take the first parabola y square is equal to 4 a x minus c1 Okay, so Let's differentiate both sides with respect to x so 2 y dy by dx will become 4 a which implies dy by dx is going to be 2 a by y Please remember we are finding this at a point h comma k So we have to substitute the point in our dy by dx expression which simply becomes 2 a by k Okay Now let's do the same for the second curve as well, which is x square is equal to 4 a y minus c2 So the derivative of this with respect to x will be 2x and we have 4 a dy by dx coming from the right hand side So dy by dx becomes x by 2 a and again since we are finding it as at h comma k Let me replace my x with h Okay, as we already discussed in the roadmap to that dy by dx for both the parabolas should be the same Right for both the parabola should be the same So we'll equate these two. Let me call this as one. Let me call this as two Okay, so equating equating one and two We obtain H by 2 a is equal to 2 a by k which implies H k is equal to 4 a square Now having done that We will do the next step, which is the process of generalization generalization means you replace your h with x and K with y that's the typical standard procedure that we follow in case of locus questions So my answer would be x y is equal to 4 a square Which is option number C. So option number C is correct. So friends in this problem. We learned That if two curves are touching each other at a given point Then at that point the slope of the tangent drawn to both the curves should be the same and Secondly, we learned how to actually deal with a locus problem. So keep practicing locus problems