the value of `lambda` for which the line `2x-8/3lambday=-3` is a normal to the conic `x^2+y^2/4=1` is:

Options

A. `(3)/(8)`
B. `(1)/(2)`
C. `-(sqrt(3))/(2)`
D. `(sqrt(3))/(2)`

Explanation

To solve the problem, we need to find the value of \( \lambda \) for which the line

\[ 2x - \frac{8}{3\lambda}y = -3 \]

is a normal to the conic

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

### Step 1: Rewrite the equation of the line

First, we can rearrange the line equation into the standard form \( Ax + By + C = 0 \):

\[ 2x + \frac{8}{3\lambda}y + 3 = 0. \]

### Step 2: Identify the conic and its properties

The given conic is an ellipse represented by

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

From this, we can identify that \( a^2 = 1 \) and \( b^2 = 4 \), which gives us \( a = 1 \) and \( b = 2 \).

### Step 3: Equation of the normal to the ellipse

The equation of the normal to the ellipse at a point defined by the angle \( \theta \) is given by:

\[ \frac{x}{\cos \theta} - \frac{2y}{\sin \theta} = 1 - 4. \]

This simplifies to:

\[ \frac{x}{\cos \theta} - \frac{2y}{\sin \theta} = -3. \]

### Step 4: Comparing coefficients

Now, we compare the coefficients of the normal equation with the line equation. From the normal equation, we can express it as:

\[ x + 2y \cdot \frac{1}{\sin \theta} = -3 \cos \theta. \]

Comparing this with \( 2x + \frac{8}{3\lambda}y + 3 = 0 \), we can equate the coefficients:

1. For \( x \): \( \frac{1}{\cos \theta} = 2 \) implies \( \cos \theta = \frac{1}{2} \).
2. For \( y \): \( -\frac{2}{\sin \theta} = \frac{8}{3\lambda} \).

### Step 5: Finding \( \sin \theta \)

From \( \cos \theta = \frac{1}{2} \), we know that \( \theta = 60^\circ \) or \( \theta = 300^\circ \). Thus, \( \sin \theta = \frac{\sqrt{3}}{2} \).

### Step 6: Substitute \( \sin \theta \) into the equation

Now substituting \( \sin \theta \) into the equation for \( y \):

\[ -\frac{2}{\frac{\sqrt{3}}{2}} = \frac{8}{3\lambda}. \]

This simplifies to:

\[ -\frac{4}{\sqrt{3}} = \frac{8}{3\lambda}. \]

### Step 7: Solve for \( \lambda \)

Cross-multiplying gives:

\[ -4 \cdot 3\lambda = 8\sqrt{3}. \]

Thus,

\[ -12\lambda = 8\sqrt{3} \implies \lambda = -\frac{2\sqrt{3}}{3}. \]

### Final Answer

The value of \( \lambda \) for which the line is a normal to the conic is:

\[ \lambda = -\frac{2\sqrt{3}}{3}. \]