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