+28 This problem seems to be easy but im stil lgetting the wrong answer
What is the distance between the vertices of the ellipse: 64x^2 + 25y^2 – 16x – 16y – 648 = 0?
my answer is:
a^2=90
a=9.4868
but the answer is apparently 10.21
How did they come up with this question?
ALSO THANK YOU FOR THIS WEBSITE ITS LITERALLY HELPING ME UNDERSTAND EVERYTHING AND NOW IM GOING TO GRADUATE IN AUGUST
+2 This is an equation of an ellipse. To find the center of the ellipse, we can complete the square for both x and y terms. - Pizza Tower
64x^2 - 16x + 25y^2 - 16y = 648
64(x^2 - (1/4)x) + 25(y^2 - (4/5)y) = 648
64(x - (1/8))^2 + 25(y - (2/5))^2 = 648 + 1 + 4/5
64(x - (1/8))^2 + 25(y - (2/5))^2 = 1293/5
The center of the ellipse is at ((1/8), (2/5)) and the semi-major and semi-minor axes are sqrt(1293/320) and sqrt(1293/125), respectively.
+397 Take a look at the answer above, #1.
There's actually a numerical mistake on the third line, but if you correct that it will get you to the result that you are looking for.
At the completed square version of the equation, if you substitute in the y co-ordinate of the centre you can calculate the extreme values for x, (the x co-ordinates of the vertices for that direction of the axes).
Similarly if you substitute in the x co-ordinate of the centre you can calculate the extreme values for y.
It turns out that the major axis is parallel with the y-axis so it's the distance between the extreme values of y that you need.
