Let X = \ (x, y) ∈ ℤ × ℤ: x ^ 2 8 + y ^ 2 20 < 1and y ^ 2 < 5x\ ....

Given,

Equation of ellipse $\dfrac{{x}^{2}}{8} + \dfrac{{y}^{2}}{20} = 1$

And equation of parabola ${y}^{2} = 5x$

Now intersection point of ellipse and parabola will be \left(2, \sqrt{10}\right) \text{ and } \left(2, - \sqrt{10}\right)$</p><p>Now plotting the region$\dfrac{{x}^{2}}{8} + \dfrac{{y}^{2}}{20} < 1and{y}^{2} < 5x$we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/e7ac496b-07c5-4b63-96d0-c586ebc3c5ce.png" style="width: 300px; height: 229px;" /></p><p>Now randomly chosen points will be between origin and the intersecting point,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/fe39bf9a-2da7-44d0-8ecf-2e43c948cfff.png" style="width: 300px; height: 267px;" /></p><p>So, the integer points inside region are$\left\{\left(2, 1\right), \left(2, - 1\right), \left(2, 2\right), \left(2, - 2\right), \left(2, 3\right), \left(2, - 3\right), \left(2, 0\right), \left(1, 1\right), \left(1, - 1\right), \left(1, 2\right), \left(1, - 2\right), \left(1, 0\right)\right\}$</p><p>Total number of ways to select three points$= C3 12 = 220$</p><p>Required number of triangle$= 4 \times C1 7 + 9 \times C1 5 = 73</p><p>{Note: Points are taken such a way that distance between two points are multiple of2$for example if we take points on line$x = 1$we can take following points \left\{\left(1, 2\right) \text{ and } \left(1, 0\right), \left(1, 0\right) \text{ and } \left(1, - 2\right), \left(1, 1\right) \text{ and } \left(1, - 1\right), \left(1, 2\right) \text{ and } \left(1, - 2\right)\right\}$

so there will be $4$ pairs from line $x = 1$ and will chose another point from $x = 2$ line so triangle will have a positive area. Also note that height of the triangle will always be $1$}

Hence, probability will be $= \dfrac{73}{220}$