Question

If \[ y=mx+\frac{3}{m} \] is a tangent to the parabola \[ y^2=4ax \] at the point \(P(3,\beta)\), where \(\beta<0\), then the value of \[ 3m-\beta \] is:

• A. \(2a\)
• B. \(ma\)
• C. \(a\)
• D. \(\dfrac{|\beta|}{a}\)

Hint:

For the parabola \[ y^2=4ax, \] the tangent with slope \(m\) is \[ y=mx+\frac{a}{m}. \] This formula is among the most frequently used results in parabola problems.

Answer 1

The Correct Option is C

Concept: For the parabola \[ y^2=4ax, \] the equation of the tangent having slope \(m\) is \[ y=mx+\frac{a}{m}. \] This standard tangent form is extremely useful because it directly relates the slope of the tangent to the parabola. In this problem, the tangent is already given in slope form. Therefore, by comparing it with the standard tangent equation, we can determine the slope and then locate the point of contact.

Step 1: Compare with the standard tangent form. Given tangent: \[ y=mx+\frac{3}{m}. \] Standard tangent: \[ y=mx+\frac{a}{m}. \] Comparing, \[ a=3. \]

Step 2: Find the point of contact corresponding to slope \(m\). For \[ y^2=4ax, \] the point corresponding to parameter \(t\) is \[ (at^2,2at). \] Also, \[ m=\frac1t. \] Therefore the point of contact is \[ \left(\frac{a}{m^2},\frac{2a}{m}\right). \]

Step 3: Use the given \(x\)-coordinate. Given point: \[ P(3,\beta). \] Thus, \[ \frac{a}{m^2}=3. \] Since \(a=3\), \[ \frac{3}{m^2}=3. \] \[ m^2=1. \] \[ m=\pm1. \]

Step 4: Use the condition \(\beta<0\). Since \[ \beta=\frac{2a}{m}, \] and \(\beta<0\), we take \[ m=-1. \] Thus, \[ \beta=\frac{2(3)}{-1} =-6. \]

Step 5: Evaluate \(3m-\beta\). \[ 3m-\beta = 3(-1)-(-6). \] \[ =-3+6. \] \[ =3. \] Since \[ a=3, \] we obtain \[ 3m-\beta=a. \]

Step 6: Final Conclusion. \[ \boxed{3m-\beta=a} \] Hence the correct answer is \[ \boxed{\text{Option (C)}}. \]