The number of integral points (x, y) interior | vedclass

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(D) The given curve is $\sqrt{(x + 5\sqrt{2})^2 + y^2} - \sqrt{(x - 5\sqrt{2})^2 + y^2} = 10$. This represents the right branch of a hyperbola with fo

Vedclass · Accepted Answer

$18$

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(D) The given curve is $\sqrt{(x + 5\sqrt{2})^2 + y^2} - \sqrt{(x - 5\sqrt{2})^2 + y^2} = 10$. This represents the right branch of a hyperbola with foci at $(\pm 5\sqrt{2}, 0)$ and $2a = 10$,so $a = 5$. The distance between foci is $2ae = 10\sqrt{2}$,so $e = \sqrt{2}$. The equation is $\frac{x^2}{25} - \frac{y^2}{25} = 1$,i.e.,$x^2 - y^2 = 25$ for $x > 0$.For a point to have exactly one tangent drawn to a hyperbola,it must lie on the curve itself. However,the question asks for points interior to $x^2 + y^2 Testing integral points $(x, y)$ such that $x^2 + y^2 If $x = 0$,$y^2 If $x = 1$,$1 + y^2 If $x = -1$,$1 + y^2 If $x = 2$,$4 + y^2 If $x = -2$,$4 + y^2 Summing these: $6 + 4 + 4 + 2 + 2 = 18$ points.

The number of integral points $(x, y)$ interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt{(x + 5\sqrt{2})^2 + y^2} - \sqrt{(x - 5\sqrt{2})^2 + y^2} = 10$ is (where an integral point $(x, y)$ means $x, y \in \mathbb{Z}$):

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