The equation (x-7^2/64)+(y+2)^2/9=1 represents an ellipse.Which points are the vertices of the ellipse?(7, 1) and (7, −5)(7, −10) and (7, 6)(10, −2) and (4, −2)(15, −2) and (−1, −2

Answer:Vertices of the ellipse are, (15, -2) and (-1, -2)Step-by-step explanation:The equation of the ellipse is,[tex]\dfrac{\left(x-7\right)^2}{64}+\dfrac{\left(y+2\right)^2}{9}=1[/tex]The general equation of ellipse with centre as (h, k) is,[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex]Comparing the given equation with the general form, we get the centre of the given ellipse at (7, -2) and a=8, b=3The line through the foci intersects the ellipse at two points, the vertices.We know that the coordinates of the vertices when (a>b) are,[tex]=(h\pm a,k)[/tex]So, vertices of the given ellipse are,[tex]=(7+8,-2),(7-8,-2)\\\\=(15,-2),(-1,-2)[/tex]