The points on the ellipse 4x^2 + y^2 = 4 that are farthest from the point (1,0) are; Option 1: (1/3, ±4sqrt(2)3) Option 2: (-1/3, ±4/3) Option 3: (-1/3, ±4sqrt(2)3) Option 4: (1/3, ±sqrt(2)3)

The answer is '(-1/3, ±4sqrt(2)3)'. View full solution:

CUET Mathematics - The points on the ellipse 4x^2 + y^2 = 4 that are farthest from the point (1,0) are; | PYQs + Solutions

CUET Mathematics 2026 21st May Shift 1

Calculus

Hard

applied

$\left(\frac{1}{3}, \pm\frac{4\sqrt{2}}{3}\right)$

$\left(\frac{-1}{3}, \pm\frac{4}{3}\right)$

$\left(\frac{-1}{3}, \pm\frac{4\sqrt{2}}{3}\right)$

$\left(\frac{1}{3}, \pm\frac{\sqrt{2}}{3}\right)$

✅ Correct Option: 3

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Match List-I with List-II

List-I	List-II
(a) If $x = t^2$ and $y = t^3$ , then $\frac{d^2y}{dx^2}$ at $t = 1$	(i) $-2$
(b) If $f(x) = \sqrt{x} + 1$ , then $f''(1)$	(ii) $-1$
(c) The minimum value of $f(x) = 9x^2 + 12x + 2$ is	(iii) $\frac{3}{4}$
(d) The point of inflexion of the function $f(x) = (x-2)^4 (x+1)^3$ is	(iv) $-\frac{1}{4}$

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