1. **State the problem:** Find the points on the ellipse $$64x^2 + y^2 = 64$$ that are farthest from the point $ (1,0) $. 2. **Rewrite the ellipse equation:** Divide both sides by 64 to get the standard form: $$\frac{x^2}{1} + \frac{y^2}{64} = 1$$ 3. **Distance formula:** The distance squared from a point $ (x,y) $ on the ellipse to $ (1,0) $ is $$D^2 = (x-1)^2 + (y-0)^2 = (x-1)^2 + y^2$$ 4. **Use the ellipse constraint:** From the ellipse equation, $$y^2 = 64 - 64x^2$$ 5. **Substitute into distance squared:** $$D^2 = (x-1)^2 + 64 - 64x^2 = x^2 - 2x + 1 + 64 - 64x^2 = -63x^2 - 2x + 65$$ 6. **Maximize $D^2$ with respect to $x$:** Take derivative and set to zero: $$\frac{dD^2}{dx} = -126x - 2 = 0 \implies -126x = 2 \implies x = -\frac{1}{63}$$ 7. **Check endpoints and critical point:** Since ellipse $x$ ranges between $-1$ and $1$, check $x = -1, x = 1$, and $x = -\frac{1}{63}$. - At $x = -1$: $$y^2 = 64 - 64(1) = 0 \implies y=0$$ $$D^2 = (-1 -1)^2 + 0 = 4$$ - At $x = 1$: $$y^2 = 64 - 64(1) = 0 \implies y=0$$ $$D^2 = (1-1)^2 + 0 = 0$$ - At $x = -\frac{1}{63}$: $$y^2 = 64 - 64\left(-\frac{1}{63}\right)^2 = 64 - 64\frac{1}{3969} = 64 - \frac{64}{3969} = \frac{253,952}{3969}$$ $$y = \pm \sqrt{\frac{253,952}{3969}}$$ Calculate $D^2$: $$D^2 = -63\left(-\frac{1}{63}\right)^2 - 2\left(-\frac{1}{63}\right) + 65 = -63\frac{1}{3969} + \frac{2}{63} + 65 = -\frac{63}{3969} + \frac{2}{63} + 65$$ Simplify: $$-\frac{63}{3969} = -\frac{1}{63}, \quad \frac{2}{63} - \frac{1}{63} = \frac{1}{63}$$ So, $$D^2 = \frac{1}{63} + 65 = \frac{1}{63} + \frac{4095}{63} = \frac{4096}{63} \approx 65.02$$ 8. **Conclusion:** The farthest points are at $$\left(-\frac{1}{63}, \pm \sqrt{64 - 64\left(-\frac{1}{63}\right)^2}\right) = \left(-\frac{1}{63}, \pm \frac{\sqrt{253,952}}{63}\right)$$ These points are farthest from $ (1,0) $ on the ellipse. **Final answer:** $$\boxed{\left(-\frac{1}{63}, \pm \frac{\sqrt{253,952}}{63}\right)}$$