The tangent at the point `P(x_1, y_1)` to the parabola `y^2 = 4 a x` meets the parabola `y^2 = 4 a (x + b)` at Q and R. the coordinates of the mid-point of QR are

Options

A. `(x_1-a,y_1+b)`
B. `(x_1,y_1)`
C. `(x_1+b,y_1+a)`
D. `(x_1-b,y_1-b)`

Answer

To find the coordinates of the midpoint of points Q and R where the tangent at point P(x₁, y₁) to the parabola \(y^2 = 4ax\) meets the parabola \(y^2 = 4a(x + b)\), we can follow these steps:

### Step 1: Write the equation of the tangent at P(x₁, y₁)
The equation of the tangent to the parabola \(y^2 = 4ax\) at the point \(P(x₁, y₁)\) is given by:
\[ yy₁ = 2a(x + x₁) \]
This can be rearranged to:
\[ yy₁ - 2ax - 2ax₁ = 0 \]

### Step 2: Substitute the tangent equation into the second parabola
Next, we substitute this tangent equation into the second parabola \(y^2 = 4a(x + b)\). The equation of the second parabola can be rearranged as:
\[ y^2 - 4a(x + b) = 0 \]
Substituting \(x\) from the tangent equation into this equation, we have:
\[ y^2 - 4a\left(\frac{yy₁ - 2ax₁}{2a} - b\right) = 0 \]
This simplifies to:
\[ y^2 - 2yy₁ + 4ax₁ + 4ab = 0 \]

### Step 3: Solve for y
This is a quadratic equation in \(y\):
\[ y^2 - 2yy₁ + (4ax₁ + 4ab) = 0 \]
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can find the roots \(y_Q\) and \(y_R\):
\[ y = \frac{2y₁ \pm \sqrt{(2y₁)^2 - 4(1)(4ax₁ + 4ab)}}{2} \]
This simplifies to:
\[ y = y₁ \pm \sqrt{y₁^2 - 4ax₁ - 4ab} \]

### Step 4: Find the midpoint of Q and R
The coordinates of points Q and R are:
- \(Q(x_Q, y_Q)\)
- \(R(x_R, y_R)\)

The x-coordinates of Q and R can be found from the tangent equation. Since both points lie on the same tangent line, the x-coordinates can be expressed in terms of the midpoint \(h\):
\[ h = \frac{x_Q + x_R}{2} = x₁ \]

The y-coordinates of Q and R are:
\[ k = \frac{y_Q + y_R}{2} = y₁ \]

### Final Result
Thus, the coordinates of the midpoint of QR are:
\[ (h, k) = (x₁, y₁) \]