Question

Let P(a, b) be a point on the parabola y² = 8x such that the tangent at P passes through the centre of the circle x² + y² – 10x – 14y + 65 = 0. Let A be the product of all possible values of a and B be the product of all possible values of b. Then the value of A + B is equal to

• A. 0
• B. 25
• C. 40
• D. 65

Answer 1

The Correct Option is D

To solve this problem, we need to find the conditions under which the tangent at a point \( P(a, b) \) on the parabola \( y^2 = 8x \) passes through the center of the given circle. Let's solve this step by step.

1. First, let's determine the center of the circle given by the equation \( x^2 + y^2 - 10x - 14y + 65 = 0 \).

We can rewrite the circle equation in standard form:

1. Complete the square:
   • For \( x^2 - 10x \), add and subtract \( \left(\frac{10}{2}\right)^2 = 25 \).
   • For \( y^2 - 14y \), add and subtract \( \left(\frac{14}{2}\right)^2 = 49 \).
2. Thus, the circle equation becomes:

\((x^2 - 10x + 25) + (y^2 - 14y + 49) = 25 + 49 - 65\)

1. This simplifies to:

\((x - 5)^2 + (y - 7)^2 = 9\)

The center of this circle is \( (5, 7) \).

1. Next, find the equation of the tangent line at \( P(a, b) \). For a parabola \( y^2 = 8x \), the tangent at point \( (a, b) \) is given by:

\(y \cdot b = 4(x + a)\)

Re-writing the tangent equation in the form \( y = mx + c \):

\(y = \frac{4}{b}x + \frac{4a}{b}\)

1. Since the tangent passes through the center of the circle \( (5, 7) \), substitute these values into the tangent line equation:

\(7 = \frac{4}{b} \cdot 5 + \frac{4a}{b}\)

1. Solving this equation for \( a \) and \( b \):

\(7b = 20 + 4a\)

\(4a = 7b - 20\)

1. Since \( b^2 = 8a \) (because point \( (a, b) \) lies on the parabola \( y^2 = 8x \)), substitute for \( a \) in terms of \( b \):

\(b^2 = 8\left(\frac{7b - 20}{4}\right)\)

\(b^2 = 2(7b - 20)\)

\(b^2 = 14b - 40\)

\(b^2 - 14b + 40 = 0\)

The possible solutions for \( b \) are the roots of this quadratic equation. Using the quadratic formula:

1. Find \( b \):

\(b = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}\)

\(b = \frac{14 \pm \sqrt{196 - 160}}{2}\)

\(b = \frac{14 \pm \sqrt{36}}{2}\)

\(b = \frac{14 \pm 6}{2}\)

\(b = 10 \quad \text{or} \quad b = 4\)

1. Calculate values of \( a \) for each \( b \):
   • If \( b = 10 \), then \( a = \frac{b^2}{8} = \frac{100}{8} = 12.5 \)
   • If \( b = 4 \), then \( a = \frac{b^2}{8} = \frac{16}{8} = 2 \)
2. Compute the products:
   • Product of all possible values of \( a \): \( A = 12.5 \times 2 = 25 \)
   • Product of all possible values of \( b \): \( B = 10 \times 4 = 40 \)
3. Final result:

\(A + B = 25 + 40 = 65\)

The value of \( A + B \) is 65, matching the correct option.