Sir,
Can the following be solved..
Find the equation of an ellipse centered at (9,6) with a Major axis of length 6 parallel to y axis.
Solution:
The closest I can get is((x-9)^2)/b^2 + ((y-6)^2)/9 =1..I've tried but still can't get an further.
Can you advise,
I somehow missed this yesterday. Very sorry. I’ll also be posting it with today’s work.
You have the correct answer except that there is a restriction on b, which within this restriction is arbitrary.
An ellipse centered at (9, 6) has the form (x-9)^2 / a^2 + (y-6)^2 / b^2 = 1. a and b will be the lengths of the semi-axes in the x and y directions.
If the axis in the y direction has length 6 then the semiaxis has length 3, so b = 3 and the equation becomes ((x-9)^2)/b^2 + ((y-6)^2)/9 =1, as you say.
You aren’t given enough information to find b, but since the major axis is in the y direction you know that b < 3. So the final answer would be simply
((x-9)^2)/b^2 + ((y-6)^2)/9 =1, b < 3.