`x-2y+4=0` is a common tangent to `y^2=4x and x^4/4+y^2/b^2=1.` Then the value of b and the other common tangent are given by : (A) `b=sqrt3` (B) `x+2y+4=0` (C) `b=3` (D) `x-2y-4=0`

Options

A. `b=sqrt3,x+2y+4=0`
B. `b=3,x+2y+4=0`
C. `b=sqrt3,x+2y-4=0`
D. `b=sqrt3,x-2y-4=0`

Answer

To solve the problem, we need to find the value of \( b \) and the other common tangent to the given parabola and ellipse. Let's break down the solution step by step.

### Step 1: Identify the equations
The given equations are:
1. The parabola: \( y^2 = 4x \)
2. The ellipse: \( \frac{x^2}{4} + \frac{y^2}{b^2} = 1 \)
3. The common tangent: \( x - 2y + 4 = 0 \)

### Step 2: Find the slope and y-intercept of the common tangent
The equation of the common tangent can be rewritten in slope-intercept form:
\[ x - 2y + 4 = 0 \implies 2y = x + 4 \implies y = \frac{1}{2}x + 2 \]
From this, we can see that the slope \( m \) of the tangent line is \( \frac{1}{2} \).

### Step 3: Use the slope to find the equation of the tangent to the ellipse
The general equation of the tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Here, \( a^2 = 4 \) (since \( a = 2 \)) and \( m = \frac{1}{2} \).

### Step 4: Substitute values into the tangent equation
Substituting \( m = \frac{1}{2} \) into the tangent equation:
\[ y = \frac{1}{2}x \pm \sqrt{4 \cdot \left(\frac{1}{2}\right)^2 + b^2} \]
Calculating \( 4 \cdot \left(\frac{1}{2}\right)^2 \):
\[ 4 \cdot \frac{1}{4} = 1 \]
Thus, the equation becomes:
\[ y = \frac{1}{2}x \pm \sqrt{1 + b^2} \]

### Step 5: Set the y-intercept equal to the common tangent
The y-intercept of the common tangent \( x - 2y + 4 = 0 \) is \( 2 \). Therefore, we set:
\[ 2 = \sqrt{1 + b^2} \]

### Step 6: Solve for \( b \)
Squaring both sides:
\[ 4 = 1 + b^2 \implies b^2 = 3 \implies b = \sqrt{3} \]

### Step 7: Find the other common tangent
Now we need to find the other common tangent using the negative slope:
Using \( m = -\frac{1}{2} \):
\[ y = -\frac{1}{2}x \pm \sqrt{4 \cdot \left(-\frac{1}{2}\right)^2 + b^2} \]
Calculating:
\[ 4 \cdot \left(-\frac{1}{2}\right)^2 = 1 \]
Thus, we have:
\[ y = -\frac{1}{2}x \pm \sqrt{1 + 3} = -\frac{1}{2}x \pm 2 \]
This gives us two equations:
1. \( y = -\frac{1}{2}x + 2 \)
2. \( y = -\frac{1}{2}x - 2 \)

Rearranging the first equation:
\[ \frac{1}{2}x + y - 2 = 0 \implies x + 2y - 4 = 0 \]

Thus, the other common tangent is:
\[ x + 2y + 4 = 0 \]

### Final Answers
- The value of \( b \) is \( \sqrt{3} \).
- The other common tangent is \( x + 2y + 4 = 0 \).