Question

Let a common tangent to the curves y² = 4x and (x – 4)² + y² = 16 touch the curves at the points P and Q. Then (PQ)² is equal to ________.

Answer 1

Correct Answer: 32

Given: \(y^2 = 4x\), \( (x-4)^2 + y^2 = 16  .\)

Step 1: Equation of the tangent to the parabola: The equation of the tangent to the parabola \( y^2 = 4x \) is:

\( y = mx + \frac{1}{m} \) ... (1)

This tangent also touches the circle \( (x - 4)^2 + y^2 = 16 \).

Step 2: Condition for tangency: The perpendicular distance from the center of the circle (4,0) to the tangent \( y = mx + \frac{1}{m} \) is equal to the radius of the circle:

\( \frac{\left| 4m + \frac{1}{m} \right|}{\sqrt{1 + m^2}} = 4 \)

Step 3: Solve for m: Square both sides:

\( \left( \frac{4m + \frac{1}{m}}{\sqrt{1 + m^2}} \right)^2 = 16 \)

\( \frac{ \left(4m + \frac{1}{m} \right)^2 }{1 + m^2} = 16 \)

\( \left( 4m + \frac{1}{m} \right)^2 = 16(1 + m^2) \)

\( 16m^2 + 8m + \frac{1}{m^2} = 16 + 16m^2 \)

\( 8m + \frac{1}{m^2} = 16 \)

Multiply through by \( m^2 \):

\( 8m^3 + 1 = 16m^2 \)

\( 8m^3 - 16m^2 + 1 = 0 \)

Solve for distinct \( m^2 \):

\( m^2 = \frac{1}{8} \)

Step 4: Point of contact of the parabola: For \( m^2 = \frac{1}{8} \), substitute into \( y = mx + \frac{1}{m} \):

\( P \left( 8, 4 \sqrt{2} \right) \)

Step 5: Calculate \( (PQ)^2 \): The distance between P and Q is given by:

\( PQ = \sqrt{S_1} \Rightarrow (PQ)^2 = S_1 \)

\( S_1 = 16 + 32 - 16 = 32 \)

Final Answer:

\( (PQ)^2 = 32 \)