1. Write the curve in standard form

The curve is $4x^2+5y^2=20$ Dividing by $20$, $\frac{x^2}{5}+\frac{y^2}{4}=1$ This is an ellipse centered at the origin.

We need the point $P(x,y)$ on this ellipse that is farthest from $Q(0,-4).$

So we maximize the squared distance: $PQ^2=(x-0)^2+(y+4)^2=x^2+(y+4)^2.$

2. Use the constraint to eliminate one variable

From the ellipse equation, $4x^2+5y^2=20 \implies x^2=\frac{20-5y^2}{4}.$

Substitute into $PQ^2$: $PQ^2=\frac{20-5y^2}{4}+(y+4)^2.$

Now simplify: $PQ^2=5-\frac{5}{4}y^2+y^2+8y+16$ $=21+8y-\frac{1}{4}y^2.$

So we must maximize $f(y)=21+8y-\frac{1}{4}y^2,$ with $y\in[-2,2]$ on the ellipse.

3. Differentiate and find the critical point

$f'(y)=8-\frac{1}{2}y.$ Setting this to zero, $8-\frac{1}{2}y=0 \implies y=16.$

But this is not possible since on the ellipse $y\in[-2,2]$.

Also, $f''(y)=-\frac{1}{2}<0,$ so $f(y)$ is concave downward. Since the vertex lies at $y=16$ (outside the allowed interval), $f(y)$ is increasing throughout $[-2,2]$.

Hence the maximum occurs at the largest allowed value: $y=2.$

4. Find the corresponding point

Using the ellipse equation: $4x^2+5(2)^2=20$ $4x^2+20=20$ $x^2=0 \implies x=0.$

So the farthest point is $P=(0,2).$

5. Compute $PQ^2$

$PQ^2=(0-0)^2+(2-(-4))^2=6^2=36.$

6. Check options

• A: $36$ ✅
• B: $48$ ❌
• C: $21$ ❌
• D: $29$ ❌

Therefore, the correct answer is A.