The ellipse `x^2/a^2+y^2/b^2=1,(a gt b)` passes through `(2,3)` and have eccentricity equal to `1/2`. Then find the equation of normal to the ellipse at `(2,3)`.

Answer

To find the equation of the normal to the ellipse at the point (2, 3), we will follow these steps:

### Step 1: Use the given information to find the values of \(a\) and \(b\).

The equation of the ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

It passes through the point (2, 3), so we substitute \(x = 2\) and \(y = 3\):

\[ \frac{2^2}{a^2} + \frac{3^2}{b^2} = 1 \]

This simplifies to:

\[ \frac{4}{a^2} + \frac{9}{b^2} = 1 \]

### Step 2: Use the eccentricity to find a relationship between \(a\) and \(b\).

The eccentricity \(e\) of the ellipse is given as \( \frac{1}{2} \). The relationship between \(a\), \(b\), and \(e\) is:

\[ b^2 = a^2(1 - e^2) \]

Substituting \(e = \frac{1}{2}\):

\[ b^2 = a^2\left(1 - \left(\frac{1}{2}\right)^2\right) = a^2\left(1 - \frac{1}{4}\right) = a^2\left(\frac{3}{4}\right) \]

### Step 3: Substitute \(b^2\) into the ellipse equation.

Substituting \(b^2 = \frac{3}{4}a^2\) into the ellipse equation we derived earlier:

\[ \frac{4}{a^2} + \frac{9}{\frac{3}{4}a^2} = 1 \]

This simplifies to:

\[ \frac{4}{a^2} + \frac{12}{a^2} = 1 \]

Combining the fractions:

\[ \frac{16}{a^2} = 1 \]

Thus, we find:

\[ a^2 = 16 \quad \Rightarrow \quad a = 4 \]

### Step 4: Find \(b^2\).

Using \(b^2 = \frac{3}{4}a^2\):

\[ b^2 = \frac{3}{4} \cdot 16 = 12 \quad \Rightarrow \quad b = 2\sqrt{3} \]

### Step 5: Write the equation of the ellipse.

Now we can write the equation of the ellipse:

\[ \frac{x^2}{16} + \frac{y^2}{12} = 1 \]

### Step 6: Differentiate to find the slope of the tangent.

To find the slope of the normal, we first need the slope of the tangent. We differentiate the equation of the ellipse implicitly:

\[ \frac{d}{dx}\left(\frac{x^2}{16} + \frac{y^2}{12}\right) = 0 \]

This gives:

\[ \frac{2x}{16} + \frac{2y}{12}\frac{dy}{dx} = 0 \]

Simplifying, we get:

\[ \frac{x}{8} + \frac{y}{6}\frac{dy}{dx} = 0 \]

Rearranging for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = -\frac{6x}{8y} = -\frac{3x}{4y} \]

### Step 7: Evaluate the slope at the point (2, 3).

Substituting \(x = 2\) and \(y = 3\):

\[ \frac{dy}{dx} = -\frac{3 \cdot 2}{4 \cdot 3} = -\frac{6}{12} = -\frac{1}{2} \]

### Step 8: Find the slope of the normal.

The slope of the normal line is the negative reciprocal of the slope of the tangent:

\[ m_{\text{normal}} = -\frac{1}{\left(-\frac{1}{2}\right)} = 2 \]

### Step 9: Write the equation of the normal line.

Using the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

Substituting \(m = 2\), \(x_1 = 2\), and \(y_1 = 3\):

\[ y - 3 = 2(x - 2) \]

Expanding this gives:

\[ y - 3 = 2x - 4 \quad \Rightarrow \quad 2x - y = -1 \quad \Rightarrow \quad 2x - y + 1 = 0 \]

Thus, the equation of the normal to the ellipse at the point (2, 3) is:

\[ \boxed{2x - y + 1 = 0} \]