How do I work out the second quadrant parametrisation of 5x^2+16y^2=45.
I have x=3 cos t, y=3sqrt5/4 sin t
so far.
Thanks
Your parametric equations are right.
The second quadrant is pi/2 <= t <= pi
To find the parametric equations for the second quadrant of the ellipse given by the equation 5x^2 + 16y^2 = 45, follow these steps:
Step 1: Write the equation of the ellipse in standard form.
Rearrange the equation to get the x^2 and y^2 terms on the left side:
5x^2 + 16y^2 = 45.
Divide both sides by 45:
x^2/9 + y^2/(45/16) = 1.
Rewrite the fractions with common denominators:
x^2/9 + y^2/[(45*16)/(16*16)] = 1.
Simplify the expression:
x^2/9 + y^2/80 = 1.
Step 2: Identify the values of a and b.
The equation is in the form x^2/a^2 + y^2/b^2 = 1.
From the standard form, we can determine that a = 3 and b = √(80)/4.
Step 3: Find the values of x and y using parametric equations.
In the second quadrant, both x and y will be negative. Therefore, adjust the signs accordingly:
x = -3cos(t)
y = -(√(80)/4)sin(t).
Thus, the second quadrant parametric equations for the ellipse 5x^2 + 16y^2 = 45 are:
x = -3cos(t)
y = -(√(80)/4)sin(t).
Make sure to specify the range of t for which you want to generate the parameterization of the ellipse. The common range used is 0 ≤ t ≤ π, but it can vary depending on the context of the problem.