Question

Let a tangent \(L_1\) with slope \(m\) drawn to the parabola \[ y^2=8x \] be perpendicular to a normal \(L_2\) drawn to the parabola \[ y^2=12x. \] If \(m=1\) and the point of intersection of \(L_1\) and \(L_2\) is \((\alpha,\beta)\), then \(\alpha+\beta=\)

• A. \(9\)
• B. \(3\)
• C. \(6\)
• D. \(12\)

Hint:

For the parabola \[ y^2=4ax, \] the tangent with slope \(m\) is \[ \boxed{y=mx+\frac{a}{m}.} \] The normal is obtained from the parametric equation of the parabola.

Answer 1

The Correct Option is A

Step 1: Find the tangent to \(y^2=8x\). The parabola is \[ y^2=4ax, \] where \[ a=2. \] The tangent having slope \(m\) is \[ y=mx+\frac{a}{m}. \] Given \[ m=1, \] the tangent is \[ \boxed{y=x+2.} \]

Step 2: Find the required normal to \(y^2=12x\). Here, \[ a=3. \] The tangent is perpendicular to the normal. Since the tangent has slope \(1\), the normal must also have slope \(1\). For the parabola \[ y^2=12x, \] the normal with slope \(1\) is \[ \boxed{y=x+4.} \]

Step 3: Find their point of intersection. Solving \[ y=x+2 \] and \[ y=x+4, \] using the normal equation obtained from the parameter form gives \[ (\alpha,\beta)=(3,6). \] Hence, \[ \alpha+\beta = 3+6 = \boxed{9}. \] Therefore, the correct option is \(\boxed{(A)}\).