Question

Let the tangent to the curve x² + 2x – 4y + 9 = 0 at the point P(1, 3) on it meet the y-axis at A. Let the line passing through P and parallel to the line x – 3y = 6 meet the parabola y² = 4x at B. If B lies on the line 2x – 3y = 8. then (AB)² is equal to___

Answer 1

Correct Answer: 292

Parabola Tangent and Intersection Problem 

The given curve is \( x^2 + 2x - 4y + 9 = 0 \). Completing the square, we get

\( (x+1)^2 = 4(y-2) \)

This is a parabola with vertex (-1, 2).

The equation of the tangent to the parabola \( (x-\alpha)^2 = 4a(y-\beta) \) at \( (x_1, y_1) \) is given by

\( (x-\alpha)(x_1 - \alpha) = 2a(y-\beta + y_1 - \beta) \)

So, the equation of the tangent at \( P(1,3) \) is

\( (x+1)(1+1) = 2 \cdot 1 (y-2+3-2) \implies 2(x+1) = 2(y-1) \implies x - y + 2 = 0 \)

This tangent meets the y-axis at A. So, the coordinates of A are (0, 2).

The line passing through \( P(1, 3) \) and parallel to \( x - 3y = 6 \) is given by

\( x - 3y + k = 0 \implies 1 - 3(3) + k = 0 \implies k = 8 \)

So the equation of the line is \( x - 3y + 8 = 0 \).

Given there appears to be an error stating that parabola is \( y^2 = 4x\) when in fact the original is \( (x+1)^2 = 4(y-2) \), let us assume that another curve intersects with the previous tangential point to be solved accordingly.

To proceed, we will continue what would of happened: This line intersects the parabola \( y^2 = 4x \). Substituting \( x = 3y - 8 \) in \( y^2 = 4x \), we have

\( y^2 = 4(3y - 8) \implies y^2 - 12y + 32 = 0 \implies (y-4)(y-8) = 0 \)

So, \( y = 4 \) or \( y = 8 \).

If \( y = 4 \), \( x = 3(4) - 8 = 4 \). If \( y = 8 \), \( x = 3(8) - 8 = 16 \).

The points of intersection are (4, 4) and (16, 8).

We are given that B lies on the line \( 2x - 3y = 8 \).

For (4, 4): \( 2(4) - 3(4) = -4 \neq 8 \). For (16, 8): \( 2(16) - 3(8) = 32 - 24 = 8 \). So, B is (16, 8).

Now, A is (0, 2) and B is (16, 8).

\( (AB)^2 = (16-0)^2 + (8-2)^2 = 16^2 + 6^2 = 256 + 36 = 292 \)

Since there may be a mistake from the original problem, will make notes to be fixed here

Conclusion:

\( \mathbf{(AB)^2 = 292} \)