Let `x=4` be a directrix to an ellipse whose centre is at the origin and its eccentricity is `(1)/(2)`. If `P(1, beta), beta gt 0` is a point on this ellipse, then the equation of the normal to it at P is :

Options

A. `7x-4y=1`
B. `4x-2y=1`
C. `8x-2y=5`
D. `4x-3y=2`

Answer

To solve the problem step by step, we need to find the equation of the normal to the ellipse at the point \( P(1, \beta) \) where \( \beta > 0 \). The ellipse has its center at the origin, a directrix at \( x = 4 \), and an eccentricity of \( \frac{1}{2} \).

### Step 1: Determine the value of \( a \)

The equation of the directrix for an ellipse is given by:

\[ x = \frac{a}{e} \]

where \( a \) is the semi-major axis and \( e \) is the eccentricity. Given that the directrix is \( x = 4 \) and \( e = \frac{1}{2} \):

\[ 4 = \frac{a}{\frac{1}{2}} \implies a = 4 \cdot \frac{1}{2} = 2 \]

### Step 2: Determine the value of \( b \)

The eccentricity \( e \) is also defined as:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the known values:

\[ \frac{1}{2} = \sqrt{1 - \frac{b^2}{2^2}} \implies \frac{1}{2} = \sqrt{1 - \frac{b^2}{4}} \]

Squaring both sides:

\[ \frac{1}{4} = 1 - \frac{b^2}{4} \]

Rearranging gives:

\[ \frac{b^2}{4} = 1 - \frac{1}{4} = \frac{3}{4} \implies b^2 = 3 \implies b = \sqrt{3} \]

### Step 3: Write the equation of the ellipse

The standard form of the ellipse centered at the origin is given by:

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

Substituting \( a^2 = 4 \) and \( b^2 = 3 \):

\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

### Step 4: Find \( \beta \) for the point \( P(1, \beta) \)

Substituting \( x = 1 \) into the ellipse equation to find \( \beta \):

\[ \frac{1^2}{4} + \frac{\beta^2}{3} = 1 \implies \frac{1}{4} + \frac{\beta^2}{3} = 1 \]

Multiplying through by 12 to eliminate the denominators:

\[ 3 + 4\beta^2 = 12 \implies 4\beta^2 = 9 \implies \beta^2 = \frac{9}{4} \implies \beta = \frac{3}{2} \]

### Step 5: Find the equation of the normal at point \( P(1, \frac{3}{2}) \)

The formula for the equation of the normal to the ellipse at point \( (x_1, y_1) \) is:

\[ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 \]

Substituting \( a^2 = 4 \), \( b^2 = 3 \), \( x_1 = 1 \), and \( y_1 = \frac{3}{2} \):

\[ \frac{4x}{1} - \frac{3y}{\frac{3}{2}} = 4 - 3 \]

This simplifies to:

\[ 4x - 2y = 1 \]

### Final Answer

The equation of the normal to the ellipse at point \( P(1, \frac{3}{2}) \) is:

\[ 4x - 2y = 1 \]