Write the equation of the tangent line in cartesian coordinates for the given parameter $t$.
$x=8\cos t$, $y=6\sin t$, $t=\frac{\pi }{4}$

• A. $y-3\sqrt{2}=-\frac{3}{4}(x-4\sqrt{2})$
• B. $y-3\sqrt{2}=\frac{3}{4}(x-4\sqrt{2})$
• C. $y-4\sqrt{2}=-\frac{3}{4}(x-3\sqrt{2})$
• D. $y-4\sqrt{2}=\frac{3}{4}(x-3\sqrt{2})$

Answer: (A)

Verified step by step guidance

1. Step 1: Recall that the equation of a tangent line can be written in the form y - y₁ = m(x - x₁), where m is the slope of the tangent line, and (x₁, y₁) is the point of tangency. Here, the parametric equations are x = 8cos(t) and y = 6sin(t), and the parameter t is given as t = π/4.
2. Step 2: Compute the coordinates of the point of tangency (x₁, y₁) by substituting t = π/4 into the parametric equations. For x₁, substitute t = π/4 into x = 8cos(t). For y₁, substitute t = π/4 into y = 6sin(t). This gives the point (x₁, y₁).
3. Step 3: Find the slope of the tangent line, m, by calculating dy/dx. Since the equations are parametric, use the chain rule: dy/dx = (dy/dt) / (dx/dt). Compute dy/dt by differentiating y = 6sin(t) with respect to t, and compute dx/dt by differentiating x = 8cos(t) with respect to t.
4. Step 4: Substitute t = π/4 into the expressions for dy/dt and dx/dt to find the slope m = (dy/dt) / (dx/dt) at t = π/4.
5. Step 5: Write the equation of the tangent line using the point-slope form y - y₁ = m(x - x₁), where m is the slope found in Step 4 and (x₁, y₁) is the point found in Step 2.