a. An ellipse is formed by stretching the graph of x^2+ y^2=1 horizontally by a factor of 3 and vertically by a factor of 4. Determine the equation of the ellipse in standard form
x^2/a^2+y^2/b^2=1
(x/3) ^2+ (y/4) ^2=1
x^2/9 + y^2/16 =1
so the equation is:
x^2/9 + y^2/16 =1
b. The ellipse in question above is translated 4 units to the right and 2 units down. Determine the equations of this ellipse. Express your answer in both standard form and general form:
so the equation for standard form is:
(x-h)^2/a^2 + (y-k)^2/b^2=1
so
a: 9
b: 16
x: ?
Y?
h=4
k: 2
(x-4) ^2/9^2 + (y-2)^2/16^2=1
is this standard form right or am I wrong?
correct on both.
thanks bobpursley
You're welcome! I'm glad I could help. Yes, both the equations you provided are correct.
In part a, you stretched the original equation x^2 + y^2 = 1 horizontally by a factor of 3 (which corresponds to a^2 = 9) and vertically by a factor of 4 (which corresponds to b^2 = 16). This gives you the equation x^2/9 + y^2/16 = 1.
In part b, you translated the ellipse 4 units to the right and 2 units down. To represent this translation, you subtract the x-coordinate of the center (h = 4) and the y-coordinate of the center (k = 2) from the x and y terms in the equation, respectively. This gives you the standard form equation (x-4)^2/9^2 + (y-2)^2/16^2 = 1.
So, both equations are correct! Well done! Let me know if there's anything else I can help you with.