If a tangent having slope 2 of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is normal to the circle `x^(2)+y^(2)+bx+1=0`, then the vlaue of `4a^(2)+b^(2)` is equal to

Options

A. 4
B. 2
C. 16
D. 8

Answer

To solve the problem step by step, we need to analyze the given information about the ellipse and the circle.

### Step 1: Write the equation of the tangent line
Given that the slope of the tangent line is 2, we can express the equation of the tangent line in slope-intercept form:
\[ y = 2x + c \]
where \( c \) is a constant.

**Hint:** Remember that the slope-intercept form of a line is \( y = mx + c \), where \( m \) is the slope.

### Step 2: Identify the center of the circle
The equation of the circle is given as:
\[ x^2 + y^2 + bx + 1 = 0 \]
We can rewrite this in standard form by completing the square:
\[ (x + \frac{b}{2})^2 + y^2 = \frac{b^2}{4} - 1 \]
From this, we can see that the center of the circle is at:
\[ \left(-\frac{b}{2}, 0\right) \]

**Hint:** Completing the square helps to identify the center and radius of a circle.

### Step 3: Determine the condition for normality
For the line \( y = 2x + c \) to be normal to the circle, it must pass through the center of the circle. Therefore, we substitute the center coordinates into the line equation:
\[ 0 = 2\left(-\frac{b}{2}\right) + c \]
This simplifies to:
\[ 0 = -b + c \]
Thus, we find:
\[ c = b \]

**Hint:** A normal line to a circle at a point must pass through the center of the circle.

### Step 4: Substitute \( c \) back into the tangent equation
Now substituting \( c \) back into the tangent line equation gives us:
\[ y = 2x + b \]

**Hint:** Ensure that you replace \( c \) with the value you found.

### Step 5: Find the point of tangency on the ellipse
The equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting \( y = 2x + b \) into the ellipse equation:
\[ \frac{x^2}{a^2} + \frac{(2x + b)^2}{b^2} = 1 \]
Expanding this:
\[ \frac{x^2}{a^2} + \frac{4x^2 + 4bx + b^2}{b^2} = 1 \]
Combining terms:
\[ \frac{x^2}{a^2} + \frac{4x^2}{b^2} + \frac{4bx}{b^2} + \frac{b^2}{b^2} = 1 \]
This simplifies to:
\[ \left(\frac{1}{a^2} + \frac{4}{b^2}\right)x^2 + \frac{4b}{b^2}x + 1 - 1 = 0 \]
This is a quadratic equation in \( x \).

**Hint:** Ensure that you correctly substitute the expression for \( y \) into the ellipse equation.

### Step 6: Condition for tangency
For the line to be tangent to the ellipse, the discriminant of this quadratic equation must be zero:
\[ \Delta = B^2 - 4AC = 0 \]
Here, \( A = \frac{1}{a^2} + \frac{4}{b^2} \), \( B = \frac{4b}{b^2} \), and \( C = 0 \).

Setting the discriminant to zero:
\[ \left(\frac{4b}{b^2}\right)^2 - 4\left(\frac{1}{a^2} + \frac{4}{b^2}\right)(0) = 0 \]

**Hint:** The condition for tangency involves the discriminant of the quadratic equation being zero.

### Step 7: Solve for \( 4a^2 + b^2 \)
From the tangency condition, we can derive relationships between \( a \) and \( b \). After some algebra, we will find that:
\[ 4a^2 + b^2 = b^2 \]
This implies that:
\[ 4a^2 = 0 \implies a^2 = 0 \]

Thus, we conclude that:
\[ 4a^2 + b^2 = b^2 \]

**Final Answer:** The value of \( 4a^2 + b^2 \) is equal to \( b^2 \).