If the point P on the curve `4x^2+5y^2-20=0` is farthest from the point `Q(0,-4)` than Find `PQ^2`.

Answer

To solve the problem of finding the square of the distance \( PQ^2 \) between a point \( P \) on the curve \( 4x^2 + 5y^2 - 20 = 0 \) and the point \( Q(0, -4) \), we can follow these steps:

### Step 1: Identify the curve
The given equation of the curve is:
\[ 4x^2 + 5y^2 - 20 = 0 \]
We can rewrite this as:
\[ \frac{x^2}{5} + \frac{y^2}{4} = 1 \]
This represents an ellipse centered at the origin with semi-major axis \( \sqrt{5} \) along the x-axis and semi-minor axis \( 2 \) along the y-axis.

### Step 2: Parametrize the point on the ellipse
A point \( P \) on the ellipse can be expressed in parametric form as:
\[ P = (a \cos \theta, b \sin \theta) = (\sqrt{5} \cos \theta, 2 \sin \theta) \]

### Step 3: Write the coordinates of point Q
The coordinates of point \( Q \) are given as:
\[ Q(0, -4) \]

### Step 4: Calculate the distance \( PQ^2 \)
The distance \( PQ \) can be calculated using the distance formula:
\[ PQ^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \]
Substituting the coordinates of points \( P \) and \( Q \):
\[ PQ^2 = \left(\sqrt{5} \cos \theta - 0\right)^2 + \left(2 \sin \theta - (-4)\right)^2 \]
This simplifies to:
\[ PQ^2 = (\sqrt{5} \cos \theta)^2 + (2 \sin \theta + 4)^2 \]
\[ = 5 \cos^2 \theta + (2 \sin \theta + 4)^2 \]

### Step 5: Expand the equation
Now, expand the second term:
\[ (2 \sin \theta + 4)^2 = 4 \sin^2 \theta + 16 \sin \theta + 16 \]
Thus, we have:
\[ PQ^2 = 5 \cos^2 \theta + 4 \sin^2 \theta + 16 \sin \theta + 16 \]

### Step 6: Use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \)
We can express \( \cos^2 \theta \) in terms of \( \sin^2 \theta \):
\[ PQ^2 = 5(1 - \sin^2 \theta) + 4 \sin^2 \theta + 16 \sin \theta + 16 \]
\[ = 5 - 5 \sin^2 \theta + 4 \sin^2 \theta + 16 \sin \theta + 16 \]
\[ = 5 + 16 + (4 - 5) \sin^2 \theta + 16 \sin \theta \]
\[ = 21 - \sin^2 \theta + 16 \sin \theta \]

### Step 7: Differentiate to find maximum distance
To find the maximum value of \( PQ^2 \), we differentiate with respect to \( \sin \theta \):
Let \( z = PQ^2 = -\sin^2 \theta + 16 \sin \theta + 21 \).
Then,
\[ \frac{dz}{d(\sin \theta)} = -2 \sin \theta + 16 \]
Setting this equal to zero for maxima:
\[ -2 \sin \theta + 16 = 0 \implies \sin \theta = 8 \]
Since \( \sin \theta \) cannot exceed 1, we check the endpoints \( \sin \theta = 1 \) and \( \sin \theta = -1 \).

### Step 8: Evaluate at endpoints
1. For \( \sin \theta = 1 \):
\[ PQ^2 = -1 + 16(1) + 21 = 36 \]
2. For \( \sin \theta = -1 \):
\[ PQ^2 = -1 - 16 + 21 = 4 \]

### Step 9: Conclusion
The maximum value occurs at \( \sin \theta = 1 \), giving:
\[ PQ^2 = 36 \]

Thus, the final answer is:
\[ \boxed{36} \]