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 consider a point \(P(a, b)\) on the parabola \(y^2 = 8x\). We know that the tangent at this point passes through the center of the given circle \(x^2 + y^2 - 10x - 14y + 65 = 0\).

First, let's find the center of the circle. The equation of the circle can be rewritten in the standard form by completing the square:

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

Complete the square for \(x\) and \(y\):

\((x-5)^2 - 25 + (y-7)^2 - 49 = -65\)

Solving, we have:

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

So, the center of the circle is \((5, 7)\).

The equation of the parabola is \(y^2 = 8x\). For a point \(P(a, b)\) on the parabola, we have \(b^2 = 8a\).

The equation of the tangent to the parabola at \(P(a, b)\) is obtained using the point form of the tangent equation:

\(yb = 4(x + a)\) (using \(T = 0\) formula for tangents)

This tangent passes through the center of the circle \((5, 7)\), so substitute these values into the tangent equation:

\(7b = 4(5 + a)\)

Solving this, we get:

\(7b = 20 + 4a\)

Thus, \(4a = 7b - 20 \; \Longrightarrow \; a = \frac{7b - 20}{4}\)

Also, recall \(b^2 = 8a\). Substitute the value of \(a\) we obtained:

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

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

Equating, we get:

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

Rearranging gives:

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

Solving this quadratic equation using the formula \(b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\), where \(A = 1\), \(B = -14\), \(C = 40\), we find:

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

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

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

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

The possible values of \(b\) are \(b = 10\) and \(b = 4\).

For each value of \(b\), calculate \(a\) using \(a = \frac{7b - 20}{4}\):

• When \(b = 10\), \(a = \frac{7 \cdot 10 - 20}{4} = \frac{50}{4} = 12.5\)
• When \(b = 4\), \(a = \frac{7 \cdot 4 - 20}{4} = \frac{8}{4} = 2\)

The product of all possible values of \(a\) is \(A = 12.5 \times 2 = 25\).

The product of all possible values of \(b\) is \(B = 10 \times 4 = 40\).

Therefore, the required sum \(A + B = 25 + 40 = 65\).

Thus, the value of \(A + B\) is 65.

Answer 2

The correct answer is (D):
Centre of circle x² + y² – 10x –14y + 65 = 0 is at (5, 7).
Let the equation of tangent to y² = 8x is
yt = x + 2t²
which passes through (5, 7)
7t = 5 + 2t²
⇒ 2t² – 7t + 5 = 0
t = 1, \(\frac{5}{2}\)
A = 1×1²×2×(\(\frac{5}{2}\))²
= 25
B = 2×2×1×2×2×\(\frac{5}{2}\)
= 40
A+B = 65