Question

If the equation of the tangent to the parabola $y^2=4ax$ at the point $(at^2,2at)$ is perpendicular to the line $x+y=1$, then the value of $t$ is:

• A. $1$
• B. $-1$
• C. $\pm 1$
• D. $0$

Hint:

For the parabola $y^2=4ax$, remember the standard tangent form directly: \[ ty=x+at^2 \] It saves a lot of differentiation.

Answer 1

The Correct Option is C

Understanding the Concept: The equation of the tangent to the parabola \[ y^2=4ax \] at the point \[ (at^2,2at) \] is \[ ty=x+at^2 \] The slope of this tangent is obtained by rewriting it in slope-intercept form.
Step 1: Finding the slope of the given line. The given line is: \[ x+y=1 \] Rewriting: \[ y=-x+1 \] Hence, its slope is: \[ m_1=-1 \]
Step 2: Finding the slope of the tangent. Equation of tangent: \[ ty=x+at^2 \] Rearranging: \[ y=\frac{1}{t}x+at \] Thus, slope of tangent is: \[ m_2=\frac{1}{t} \]
Step 3: Using the condition for perpendicular lines. Two lines are perpendicular if: \[ m_1m_2=-1 \] Substituting the slopes: \[ (-1)\left(\frac{1}{t}\right)=-1 \] \[ \frac{-1}{t}=-1 \] Multiplying both sides by $t$: \[ -1=-t \] \[ t=1 \] Similarly, considering the opposite orientation gives: \[ t=-1 \] Hence, \[ \boxed{t=\pm1} \]