Which of the following is an exterior point of the ellipse `16x^2+9y^2-16x-32=0`?

Options

A. `(1/2,2)`
B. `(1/4,1)`
C. `(3,-2)`
D. d)none of these

Answer

To determine which of the given points is an exterior point of the ellipse defined by the equation \(16x^2 + 9y^2 - 16x - 32 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse
First, we need to rewrite the equation of the ellipse in standard form. We start with the given equation:

\[ 16x^2 + 9y^2 - 16x - 32 = 0 \]

Rearranging gives:

\[ 16x^2 - 16x + 9y^2 = 32 \]

### Step 2: Complete the square for the x-terms
Next, we complete the square for the \(x\) terms in the equation:

\[ 16(x^2 - x) + 9y^2 = 32 \]

To complete the square for \(x^2 - x\):

\[ x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4} \]

Substituting this back into the equation gives:

\[ 16\left((x - \frac{1}{2})^2 - \frac{1}{4}\right) + 9y^2 = 32 \]

Expanding this:

\[ 16(x - \frac{1}{2})^2 - 4 + 9y^2 = 32 \]

Adding 4 to both sides results in:

\[ 16(x - \frac{1}{2})^2 + 9y^2 = 36 \]

### Step 3: Divide by 36 to get the standard form
Now, we divide the entire equation by 36:

\[ \frac{16(x - \frac{1}{2})^2}{36} + \frac{9y^2}{36} = 1 \]

This simplifies to:

\[ \frac{(x - \frac{1}{2})^2}{\frac{9}{4}} + \frac{y^2}{4} = 1 \]

### Step 4: Identify the center and axes of the ellipse
From the standard form, we can identify:
- Center: \((\frac{1}{2}, 0)\)
- Semi-major axis: \(a = \frac{3}{2}\)
- Semi-minor axis: \(b = 2\)

### Step 5: Determine the points to test
Now, we will test the given points to see if they lie inside, on, or outside the ellipse. The points to test are:
1. \((\frac{1}{2}, 2)\)
2. \((\frac{1}{4}, 1)\)
3. \((3, -2)\)

### Step 6: Test the first point \((\frac{1}{2}, 2)\)
Substituting \((\frac{1}{2}, 2)\) into the original equation:

\[ 16\left(\frac{1}{2}\right)^2 + 9(2)^2 - 16\left(\frac{1}{2}\right) - 32 \]

Calculating:

\[ 16 \cdot \frac{1}{4} + 9 \cdot 4 - 8 - 32 = 4 + 36 - 8 - 32 = 0 \]

This point lies on the ellipse.

### Step 7: Test the second point \((\frac{1}{4}, 1)\)
Substituting \((\frac{1}{4}, 1)\):

\[ 16\left(\frac{1}{4}\right)^2 + 9(1)^2 - 16\left(\frac{1}{4}\right) - 32 \]

Calculating:

\[ 16 \cdot \frac{1}{16} + 9 - 4 - 32 = 1 + 9 - 4 - 32 = -26 \]

This point lies inside the ellipse.

### Step 8: Test the third point \((3, -2)\)
Substituting \((3, -2)\):

\[ 16(3)^2 + 9(-2)^2 - 16(3) - 32 \]

Calculating:

\[ 16 \cdot 9 + 9 \cdot 4 - 48 - 32 = 144 + 36 - 48 - 32 = 100 \]

This point is positive, indicating it lies outside the ellipse.

### Conclusion
The point that lies outside the ellipse is \((3, -2)\).