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The point of intersection of the tangent...

The point of intersection of the tangents at `t_(1)` and `t_(2)` to the parabola `y^(2)=12x` is

A

`[2t_(1)t_(2),2(t_(1)-t_(2))]`

B

`[3t_(1)t_(2),3(t_(1)-t_(2))]`

C

`[3t_(1)t_(2),3(t_(1)+t_(2))]`

D

`[2t_(1)t_(2),2(t_(1)+t_(2))]`

Text Solution

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To find the point of intersection of the tangents at \( t_1 \) and \( t_2 \) to the parabola given by the equation \( y^2 = 12x \), we can follow these steps: ### Step 1: Identify the equation of the parabola The equation of the parabola is given as: \[ y^2 = 12x \] This can be rewritten in the standard form \( y^2 = 4ax \), where \( 4a = 12 \). Thus, we find: \[ a = 3 \] ### Step 2: Write the equations of the tangents For a parabola \( y^2 = 4ax \), the equation of the tangent at a point \( (at^2, 2at) \) is given by: \[ ty = x + at^2 \] Using this formula, we can write the equations for the tangents at \( t_1 \) and \( t_2 \): 1. For \( t_1 \): \[ t_1 y = x + 3t_1^2 \quad \text{(Equation 1)} \] 2. For \( t_2 \): \[ t_2 y = x + 3t_2^2 \quad \text{(Equation 2)} \] ### Step 3: Solve the system of equations To find the point of intersection of these two tangents, we can subtract Equation 2 from Equation 1: \[ t_1 y - t_2 y = (x + 3t_1^2) - (x + 3t_2^2) \] This simplifies to: \[ (t_1 - t_2)y = 3(t_1^2 - t_2^2) \] Using the difference of squares, we can factor the right-hand side: \[ (t_1 - t_2)y = 3(t_1 - t_2)(t_1 + t_2) \] Assuming \( t_1 \neq t_2 \), we can divide both sides by \( t_1 - t_2 \): \[ y = 3(t_1 + t_2) \] ### Step 4: Substitute \( y \) back to find \( x \) Now, we can substitute \( y = 3(t_1 + t_2) \) back into either Equation 1 or Equation 2 to find \( x \). Let's use Equation 1: \[ t_1(3(t_1 + t_2)) = x + 3t_1^2 \] This simplifies to: \[ 3t_1(t_1 + t_2) = x + 3t_1^2 \] Rearranging gives: \[ x = 3t_1(t_1 + t_2) - 3t_1^2 \] Factoring out the common terms: \[ x = 3t_1(t_1 + t_2 - t_1) = 3t_1 t_2 \] ### Step 5: Final point of intersection Thus, the point of intersection of the tangents at \( t_1 \) and \( t_2 \) is: \[ (3t_1 t_2, 3(t_1 + t_2)) \] ### Summary The point of intersection of the tangents at \( t_1 \) and \( t_2 \) to the parabola \( y^2 = 12x \) is: \[ (3t_1 t_2, 3(t_1 + t_2)) \]
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