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The locus of centroid of triangle formed...

The locus of centroid of triangle formed by a tangent to the parabola `y^(2) = 36x` with coordinate axes is

A

`y^(2) =- 9x`

B

`y^(2) +3x = 0`

C

`y^(2) = 3x`

D

`y^(2) = 9x`

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To find the locus of the centroid of the triangle formed by a tangent to the parabola \( y^2 = 36x \) and the coordinate axes, we can follow these steps: ### Step 1: Equation of the Tangent The standard form of the parabola is given by \( y^2 = 4ax \). Here, \( 4a = 36 \), so \( a = 9 \). The equation of the tangent to the parabola at point \( (t^2, 2at) \) is given by: \[ yy_1 = 2a(x + x_1) \] Substituting \( a = 9 \) and \( x_1 = t^2 \) and \( y_1 = 2at = 18t \): \[ yy_1 = 18(x + t^2) \] This simplifies to: \[ yy_1 = 18x + 18t^2 \] ### Step 2: Finding the Points of Intersection To find the points where the tangent intersects the axes, we set \( y = 0 \) for the x-intercept: \[ 0 = 18x + 18t^2 \implies x = -t^2 \] For the y-intercept, set \( x = 0 \): \[ yy_1 = 18(0 + t^2) \implies y = 18t \] Thus, the points of intersection are \( (-t^2, 0) \) and \( (0, 18t) \). ### Step 3: Finding the Centroid The centroid \( G \) of the triangle formed by the points \( (-t^2, 0) \), \( (0, 18t) \), and the origin \( (0, 0) \) is given by: \[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \] Substituting the coordinates: \[ G\left( \frac{-t^2 + 0 + 0}{3}, \frac{0 + 18t + 0}{3} \right) = \left( -\frac{t^2}{3}, 6t \right) \] ### Step 4: Eliminating the Parameter To find the locus, we need to eliminate the parameter \( t \). From the expression for \( G \): 1. Let \( x = -\frac{t^2}{3} \) which gives \( t^2 = -3x \). 2. Substitute \( t^2 \) into the expression for \( y \): \[ y = 6t = 6\sqrt{-3x} \quad \text{(taking the positive root)} \] Thus, squaring both sides: \[ y^2 = 36(-3x) \implies y^2 = -108x \] ### Step 5: Final Locus Equation The locus of the centroid is given by the equation: \[ y^2 + 108x = 0 \] ### Conclusion The locus of the centroid of the triangle formed by the tangent to the parabola \( y^2 = 36x \) with the coordinate axes is: \[ y^2 + 108x = 0 \]

To find the locus of the centroid of the triangle formed by a tangent to the parabola \( y^2 = 36x \) and the coordinate axes, we can follow these steps: ### Step 1: Equation of the Tangent The standard form of the parabola is given by \( y^2 = 4ax \). Here, \( 4a = 36 \), so \( a = 9 \). The equation of the tangent to the parabola at point \( (t^2, 2at) \) is given by: \[ yy_1 = 2a(x + x_1) \] Substituting \( a = 9 \) and \( x_1 = t^2 \) and \( y_1 = 2at = 18t \): ...
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