the equation for an ellipse is given by ((x-3)^2 ) / (16) + ((y+2)^2 / (b^2))?

Question

find b so that the eccentricity is .75

((x-3)^2 ) / (16) + ((y+2)^2 / (b^2))= 1
find b so that the eccentricith is 0.75

Answer 1

(x - 3) ² / 16 + (y + 2) ² / b ² = 1
b ² = a ² (1 - e ²)
b ² = 16 (1 - e ²)
b ² = 16 ( 1 - 9 /16)
b ² = 16 - 9
b ² = 7
b = √ 7

Answer 2

In such case,u hav an ellipse which is not usually used.In such case,fistly,let x-3=X,and let y+2=Y
so equation is:
X^2/16+Y^2/b^2=1
now 1-e^2=b^2/a^2
as e=0.75 or 3/4.
thrfore.b^2=7 and hence b=root(7)
in this question,our assumption was not useful.However if u want to find focus,vertex etc.
then here is how to find vertex.
for a normal ellipse which we hav considered,vertex:X=0,Y=0
->x-3=0,y+2=0
->x=3,y=-2 is the required vertex.
Similarly u can find out focus.
So remeber what u hav to do in such type questions

Answer 3

Sorry, but this is not even an equation! Is there a 1 somewhere?

Answer 4

Where is the equation? I see no equals sign.