4 attempts remaining. Find the points on the ellipse $25 x^{2}+y^{2}=25$ that are farthest away from the point $(1,0)$. List them as a list of points, such as " $(1,2),(3,4) "$. List of points: Submit answer

Step 1 ：First, we need to express y in terms of x using the equation of the ellipse. The equation of the ellipse is \(25x^{2}+y^{2}=25\). Solving for y, we get \(y = \sqrt{25 - 25x^{2}}\).

Step 2 ：Next, we substitute this into the equation for the square of the distance from the point (1,0). The square of the distance is \((x-1)^{2} + y^{2}\). Substituting \(y = \sqrt{25 - 25x^{2}}\) into this equation, we get \(d_{sq} = (x-1)^{2} + 25 - 25x^{2}\).

Step 3 ：We then differentiate this equation with respect to x to find the maximum. The derivative of \(d_{sq}\) with respect to x is \(-48x - 2\).

Step 4 ：We solve the equation \(-48x - 2 = 0\) to find the x-coordinates of the points. The solution is \(x = -1/24\).

Step 5 ：We substitute the x-values into the equation for y to find the y-coordinates. Substituting \(x = -1/24\) into the equation \(y = \sqrt{25 - 25x^{2}}\), we get \(y = 25\sqrt{23}/24\).

Step 6 ：The points on the ellipse that are farthest away from the point (1,0) are \((-1/24, 25\sqrt{23}/24)\) and \((-1/24, -25\sqrt{23}/24)\). The second point is obtained by considering the negative square root in the equation for y.

Step 7 ：Final Answer: The points on the ellipse that are farthest away from the point (1,0) are \(\boxed{(-1/24, 25\sqrt{23}/24), (-1/24, -25\sqrt{23}/24)}\).