If a tangent of slope `2` of the ellipse `(x^2)/(a^2) + (y^2)/1 = 1` passes through the point `(-2,0)`, then the value of `a^2` is equal to

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Write the equation of the ellipse
The given ellipse is represented by the equation:
\[ \frac{x^2}{a^2} + \frac{y^2}{1} = 1 \]

### Step 2: Write the equation of the tangent line
The general equation of a tangent to the ellipse can be expressed in the form:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
where \(m\) is the slope of the tangent and \(b^2 = 1\) for our ellipse. Given that the slope \(m = 2\), we can substitute this into the tangent equation.

### Step 3: Substitute the values into the tangent equation
Substituting \(m = 2\) and \(b^2 = 1\) into the tangent equation:
\[ y = 2x \pm \sqrt{a^2(2^2) + 1} \]
This simplifies to:
\[ y = 2x \pm \sqrt{4a^2 + 1} \]

### Step 4: Use the point through which the tangent passes
The tangent line passes through the point \((-2, 0)\). We will substitute \(x = -2\) and \(y = 0\) into the tangent equation:
\[ 0 = 2(-2) \pm \sqrt{4a^2 + 1} \]
This simplifies to:
\[ 0 = -4 \pm \sqrt{4a^2 + 1} \]

### Step 5: Solve for \(a^2\)
From the equation above, we can separate the two cases:
1. \(0 = -4 + \sqrt{4a^2 + 1}\)
2. \(0 = -4 - \sqrt{4a^2 + 1}\) (This case is not possible since the square root cannot be negative)

Focusing on the first case:
\[ \sqrt{4a^2 + 1} = 4 \]
Now, squaring both sides:
\[ 4a^2 + 1 = 16 \]
Subtracting 1 from both sides gives:
\[ 4a^2 = 15 \]
Dividing both sides by 4 results in:
\[ a^2 = \frac{15}{4} \]

### Final Answer
Thus, the value of \(a^2\) is:
\[ \boxed{\frac{15}{4}} \]