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If the tangents at t(1) and t(2) on y^(2...

If the tangents at `t_(1)` and `t_(2)` on `y^(2)`=4ax meet on the directrix then

A

`t_(1)=t_(2)`

B

`t_(1)=-t_(2)`

C

`t_(1)t_(2)=2`

D

`t_(1)t_(2)=-1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Understand the Tangent Equation The equation of the parabola is given as \( y^2 = 4ax \). The equation of the tangent to the parabola at the parameter \( t \) is given by: \[ y \cdot t = x + at^2 \] This means for the tangents at parameters \( t_1 \) and \( t_2 \), we have: 1. Tangent at \( t_1 \): \( y \cdot t_1 = x + at_1^2 \) (Equation 1) 2. Tangent at \( t_2 \): \( y \cdot t_2 = x + at_2^2 \) (Equation 2) ### Step 2: Set Up the Equations We will subtract Equation 1 from Equation 2 to eliminate \( x \): \[ y \cdot t_2 - y \cdot t_1 = (x + at_2^2) - (x + at_1^2) \] This simplifies to: \[ y(t_2 - t_1) = a(t_2^2 - t_1^2) \] ### Step 3: Factor the Right Side The right side can be factored using the difference of squares: \[ y(t_2 - t_1) = a(t_2 - t_1)(t_2 + t_1) \] Assuming \( t_2 \neq t_1 \), we can divide both sides by \( (t_2 - t_1) \): \[ y = a(t_2 + t_1) \] ### Step 4: Find the x-coordinate Now, we can substitute \( y = a(t_2 + t_1) \) back into either Equation 1 or Equation 2 to find \( x \). Using Equation 1: \[ a(t_2 + t_1) \cdot t_1 = x + at_1^2 \] This simplifies to: \[ at_1(t_2 + t_1) = x + at_1^2 \] Rearranging gives: \[ x = at_1(t_2 + t_1) - at_1^2 \] Factoring out \( a \): \[ x = a(t_1t_2 + t_1^2 - t_1^2) = at_1t_2 \] ### Step 5: Intersection Point Thus, the point of intersection of the tangents is: \[ \left( at_1t_2, a(t_2 + t_1) \right) \] ### Step 6: Condition for Directrix The directrix of the parabola \( y^2 = 4ax \) is given by the line \( x = -a \). For the tangents to meet on the directrix, the x-coordinate of the intersection point must equal \(-a\): \[ at_1t_2 = -a \] Dividing both sides by \( a \) (assuming \( a \neq 0 \)): \[ t_1t_2 = -1 \] ### Conclusion Thus, the condition for the tangents at \( t_1 \) and \( t_2 \) to meet on the directrix is: \[ t_1t_2 = -1 \] ---
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