UPSC NDA I 2019 Mathematics Solved Paper

Concept:The intersection points of two curves are found by solving their equations simultaneously.Both given parabolas are symmetric about axes and have the vertex at the origin.Explanation:Step 1: Solve $y^2 = 4ax$ and $x^2 = 4ay$ together.Square the first equation: $y^4 = 16a^2 x^2$.From the second equation, $x^2 = 4ay$. Substitute: $y^4 = 16a^2 (4ay) = 64a^3 y$.Thus $y^4 - 64a^3 y = 0$ ⇒ $y(y^3 - 64a^3) = 0$ ⇒ $y = 0$ or $y = 4a$.Step 2: Find corresponding $x$ values.If $y = 0$, then from $y^2 = 4ax$ we get $x = 0$. So one intersection is $(0,0)$.If $y = 4a$, then $y^2 = 16a^2 = 4ax$ ⇒ $x = 4a$. So another intersection is $(4a,4a)$.Step 3: Check the line $y = x$. Both points satisfy $y = x$ (since $0=0$ and $4a=4a$).Thus the curves intersect at two points on $y = x$. No other points exist.Answer:The correct option is A: at two points on the line $y = x$.