The equation of the normal at the point P (2, 3) on the ellipse `9x^(2) + 16y^(2) = 180`, is

Options

A. `3y = 8x - 10`
B. `3y - 8x + 7 = 0`
C. `8y + 3x + 7 = 0`
D. `3x + 2y + 7 = 0`

Answer

To find the equation of the normal at the point P(2, 3) on the ellipse given by the equation \(9x^2 + 16y^2 = 180\), we can follow these steps:

### Step 1: Write down the equation of the ellipse
The equation of the ellipse is given as:
\[ 9x^2 + 16y^2 = 180 \]

### Step 2: Differentiate the equation implicitly
To find the slope of the tangent line at point P(2, 3), we need to differentiate the equation of the ellipse with respect to \(x\):
\[ \frac{d}{dx}(9x^2) + \frac{d}{dx}(16y^2) = \frac{d}{dx}(180) \]
This gives us:
\[ 18x + 32y\frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the differentiated equation to solve for \(\frac{dy}{dx}\):
\[ 32y\frac{dy}{dx} = -18x \]
\[ \frac{dy}{dx} = -\frac{18x}{32y} = -\frac{9x}{16y} \]

### Step 4: Substitute the point P(2, 3)
Now, we substitute the coordinates of point P(2, 3) into the derivative to find the slope of the tangent line:
\[ \frac{dy}{dx} = -\frac{9(2)}{16(3)} = -\frac{18}{48} = -\frac{3}{8} \]

### Step 5: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{\left(-\frac{3}{8}\right)} = \frac{8}{3} \]

### Step 6: Use the point-slope form to write the equation of the normal
Using the point-slope form of the equation of a line, we have:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = \frac{8}{3}\), \(x_1 = 2\), and \(y_1 = 3\):
\[ y - 3 = \frac{8}{3}(x - 2) \]

### Step 7: Rearranging to standard form
Now, we rearrange this equation:
\[ y - 3 = \frac{8}{3}x - \frac{16}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y - 9 = 8x - 16 \]
Rearranging gives:
\[ 8x - 3y + 7 = 0 \]

### Final Answer
The equation of the normal at the point P(2, 3) on the ellipse is:
\[ 8x - 3y + 7 = 0 \]