If tangent to the parabola `y^2=8x` at (2,-4) also touches the circle `x^2+y^2=a`. then find the value of a

Answer

To solve the problem, we need to find the value of \( a \) such that the tangent to the parabola \( y^2 = 8x \) at the point \( (2, -4) \) also touches the circle \( x^2 + y^2 = a \).

### Step 1: Find the equation of the tangent to the parabola at the given point

The equation of the parabola is given by:
\[ y^2 = 8x \]
To find the tangent line at the point \( (2, -4) \), we first need to find the slope of the tangent at that point. We differentiate the equation of the parabola with respect to \( x \).

Differentiating \( y^2 = 8x \):
\[ 2y \frac{dy}{dx} = 8 \implies \frac{dy}{dx} = \frac{8}{2y} = \frac{4}{y} \]

At the point \( (2, -4) \):
\[ \frac{dy}{dx} = \frac{4}{-4} = -1 \]

Now, using the point-slope form of the line, the equation of the tangent line at \( (x_1, y_1) = (2, -4) \) is:
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y + 4 = -1(x - 2) \]
Simplifying this:
\[ y + 4 = -x + 2 \implies x + y + 2 = 0 \]

### Step 2: Find the distance from the center of the circle to the tangent line

The equation of the circle is:
\[ x^2 + y^2 = a \]
The center of the circle is at \( (0, 0) \) and the radius is \( \sqrt{a} \).

The distance \( D \) from the center \( (0, 0) \) to the line \( x + y + 2 = 0 \) can be calculated using the formula for the distance from a point to a line:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
where \( A = 1, B = 1, C = 2 \) and \( (x_1, y_1) = (0, 0) \).

Calculating the distance:
\[ D = \frac{|1 \cdot 0 + 1 \cdot 0 + 2|}{\sqrt{1^2 + 1^2}} = \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 3: Set the distance equal to the radius

Since the tangent line touches the circle, the distance \( D \) must equal the radius \( \sqrt{a} \):
\[ \sqrt{2} = \sqrt{a} \]

### Step 4: Solve for \( a \)

Squaring both sides:
\[ 2 = a \]

Thus, the value of \( a \) is:
\[ \boxed{2} \]