UPSC NDA II 2018 Mathematics Solved Paper

Concept:For a second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, if the discriminant $\Delta = 0$ and $h^2 - ab < 0$, the equation represents a single point (degenerate ellipse).Explanation:Given: $x^2 + 4y^2 - 2x - 4y + 2 = 0$.Complete the squares:$(x^2 - 2x) + (4y^2 - 4y) + 2 = 0$$(x-1)^2 - 1 + 4[(y - \frac{1}{2})^2 - \frac{1}{4}] + 2 = 0$$(x-1)^2 - 1 + 4(y - \frac{1}{2})^2 - 1 + 2 = 0$$(x-1)^2 + 4(y - \frac{1}{2})^2 = 0$.Since both squares are non-negative, their sum is zero only when each is zero:$x-1=0$ and $y-\frac{1}{2}=0$.Thus $(x, y) = (1, \frac{1}{2})$.So the equation represents a single point.Check discriminant: $a=1, b=4, h=0, g=-1, f=-2, c=2$.$\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 1\cdot4\cdot2 + 0 - 1\cdot(-2)^2 - 4\cdot(-1)^2 = 8 - 4 - 4 = 0$.$h^2 - ab = 0 - 4 = -4 < 0$.Thus it is a degenerate ellipse (a point).Answer:A. a point