MHTCET 23 Apr 2025 Shift 2 Paper

Concept:The given equation represents an ellipse with its major axis along the $y$-axis, so the foci lie on the line $x = h$ at $(h, k \pm c)$.Explanation:Rewrite the equation by completing the square for $x$.$25x^{2} + 16y^{2} - 150x = 175$$25(x^{2} - 6x) + 16y^{2} = 175$$25[(x-3)^{2} - 9] + 16y^{2} = 175$$25(x-3)^{2} + 16y^{2} = 175 + 225 = 400$Divide throughout by $400$:$\frac{(x-3)^{2}}{16} + \frac{y^{2}}{25} = 1$Here, the larger denominator is under $y^{2}$, so $a^{2} = 25$ and $b^{2} = 16$.For an ellipse, $c^{2} = a^{2} - b^{2} = 25 - 16 = 9$, so $c = 3$.The centre is $(h, k) = (3, 0)$.Since the major axis is vertical, the foci are at $(h, k \pm c) = (3, 0 \pm 3)$.Answer:Foci are $(3, \pm 3)$, hence option B is correct.