Let common tangent to `x^2+y^2=9/4` and `y^2=4x` meets x-axis at point Q . If an ellipse with length of major axis equal to 6 and minor axis is of length `'OQ'` is drawn , then the value of `l/e^2` is

Options

A. 2
B. 5
C. 7
D. 8

Answer

To solve the problem step by step, we will follow the given information and derive the necessary values.

### Step 1: Identify the equations of the curves
The equations given are:
1. Circle: \( x^2 + y^2 = \frac{9}{4} \) (which has a radius of \( \frac{3}{2} \))
2. Parabola: \( y^2 = 4x \)

### Step 2: Find the equation of the common tangent
The equation of the common tangent to the circle can be expressed as:
\[ y = mx \pm \sqrt{\frac{9}{4}(1 + m^2)} \]

For the parabola, the equation of the tangent is:
\[ y = mx + \frac{1}{m} \]

Since these are common tangents, we can equate the two expressions for \( y \):
\[ mx + \frac{1}{m} = mx \pm \sqrt{\frac{9}{4}(1 + m^2)} \]

### Step 3: Solve for \( m \)
By equating the two expressions, we can isolate \( \frac{1}{m} \):
\[ \frac{1}{m} = \pm \sqrt{\frac{9}{4}(1 + m^2)} \]

Squaring both sides gives:
\[ \frac{1}{m^2} = \frac{9}{4}(1 + m^2) \]

Rearranging this leads to:
\[ 4 = 9m^2 + 9m^4 \]
\[ 9m^4 + 9m^2 - 4 = 0 \]

Let \( t = m^2 \). Then, the equation becomes:
\[ 9t^2 + 9t - 4 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 9, b = 9, c = -4 \):
\[ t = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 9 \cdot (-4)}}{2 \cdot 9} \]
\[ t = \frac{-9 \pm \sqrt{81 + 144}}{18} \]
\[ t = \frac{-9 \pm \sqrt{225}}{18} \]
\[ t = \frac{-9 \pm 15}{18} \]

Calculating the two possible values:
1. \( t = \frac{6}{18} = \frac{1}{3} \)
2. \( t = \frac{-24}{18} \) (not valid since \( t \) must be non-negative)

Thus, \( m^2 = \frac{1}{3} \) and \( m = \pm \frac{1}{\sqrt{3}} \).

### Step 5: Find the tangent line equation
Taking \( m = \frac{1}{\sqrt{3}} \):
\[ y = \frac{1}{\sqrt{3}}x + \sqrt{3} \]

### Step 6: Find the intersection with the x-axis
Set \( y = 0 \):
\[ 0 = \frac{1}{\sqrt{3}}x + \sqrt{3} \]
\[ \frac{1}{\sqrt{3}}x = -\sqrt{3} \]
\[ x = -3 \]

Thus, point \( Q \) is \( (-3, 0) \).

### Step 7: Calculate the length \( OQ \)
The distance \( OQ \) (from origin \( O(0,0) \) to \( Q(-3,0) \)) is:
\[ OQ = 3 \]

### Step 8: Define the ellipse
The ellipse has a major axis of length \( 6 \) and a minor axis of length \( OQ = 3 \):
- Major axis \( 2a = 6 \) → \( a = 3 \)
- Minor axis \( 2b = 3 \) → \( b = \frac{3}{2} \)

### Step 9: Calculate the eccentricity \( e \)
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
\[ b^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}, \quad a^2 = 3^2 = 9 \]
\[ e^2 = 1 - \frac{9/4}{9} = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 10: Calculate \( \frac{l}{e^2} \)
The length of the latus rectum \( l \) is given by:
\[ l = \frac{2b^2}{a} = \frac{2 \cdot \frac{9}{4}}{3} = \frac{9}{6} = \frac{3}{2} \]

Now calculate \( \frac{l}{e^2} \):
\[ \frac{l}{e^2} = \frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3}{2} \cdot \frac{4}{3} = 2 \]

### Final Answer
Thus, the value of \( \frac{l}{e^2} \) is \( 2 \).

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