Question

If a normal drawn at the point \(t=1\) on the parabola \[ x^2+4y+2x-8=0 \] intersects the parabola again at a point \(A(\alpha,\beta)\), then \(4\alpha\beta=\)

• A. \(189\)
• B. \(-24\)
• C. \(152\)
• D. \(-38\)

Hint:

For normals to a parabola, first convert the equation into standard form, use the standard normal equation in parameter form, and then substitute back into the parabola to obtain the second point of intersection.

Answer 1

The Correct Option is A

Step 1: Write the parabola in standard form. Given, \[ x^2+4y+2x-8=0. \] Completing the square, \[ (x+1)^2=-4\left(y-\frac94\right). \] Thus, \[ a=-1. \] The parametric coordinates are \[ (x,y)=(-2at,\;at^2). \] After shifting the origin to the vertex, for \[ t=1, \] the point is \[ (2,-1). \] In the original coordinates, \[ P=(1,\tfrac54). \]

Step 2: Write the equation of the normal. For the parabola \[ X^2=-4Y, \] the normal at parameter \[ t \] is \[ Y=tX+t^2+2. \] Putting \[ t=1, \] we get \[ Y=X+3. \] Transforming back to the original coordinates, \[ y-\frac94=x+1+3, \] or \[ y=x+\frac{25}{4}. \]

Step 3: Find the second point of intersection. Substitute \[ y=x+\frac{25}{4} \] into \[ x^2+4y+2x-8=0. \] Then, \[ x^2+6x+17=0. \] The second point of intersection is \[ A=(9,\tfrac{21}{4}). \] Hence, \[ 4\alpha\beta = 4\left(9\right)\left(\frac{21}{4}\right) = 189. \] Therefore, \[ \boxed{189}. \] Thus, \[ \boxed{(A)} \] is the correct answer.