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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`

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: Identify the Parabola The given parabola is \( y^2 = 36x \). This can be compared to the standard form \( y^2 = 4ax \), where \( 4a = 36 \). Thus, we find \( a = 9 \). ### Step 2: Write the Equation of the Tangent The equation of a tangent to the parabola \( y^2 = 4ax \) at the point \( (x_1, y_1) \) is given by: \[ yy_1 = 2a(x + x_1) \] Substituting \( a = 9 \), we get: \[ yy_1 = 18(x + x_1) \] ### Step 3: Find Points of Intersection with Axes 1. **Intersection with the x-axis**: Set \( y = 0 \): \[ 0 = 18(x + x_1) \implies x = -x_1 \] So the point of intersection with the x-axis is \( A(-x_1, 0) \). 2. **Intersection with the y-axis**: Set \( x = 0 \): \[ yy_1 = 18(0 + x_1) \implies y = \frac{18x_1}{y_1} \] So the point of intersection with the y-axis is \( B(0, \frac{18x_1}{y_1}) \). ### Step 4: Find the Centroid of the Triangle The vertices of the triangle are \( A(-x_1, 0) \), \( B(0, \frac{18x_1}{y_1}) \), and \( O(0, 0) \). The coordinates of the centroid \( G \) are given by: \[ G\left( \frac{x_1^1 + x_2^2 + x_3^3}{3}, \frac{y_1^1 + y_2^2 + y_3^3}{3} \right) \] Substituting the coordinates: \[ G\left( \frac{-x_1 + 0 + 0}{3}, \frac{0 + \frac{18x_1}{y_1} + 0}{3} \right) = \left(-\frac{x_1}{3}, \frac{6x_1}{y_1}\right) \] ### Step 5: Express \( y_1 \) in terms of \( x_1 \) Since \( (x_1, y_1) \) lies on the parabola \( y^2 = 36x \), we have: \[ y_1^2 = 36x_1 \] From the centroid coordinates, we can express \( y_1 \) as: \[ y_1 = \frac{6x_1}{q} \] Substituting this into the parabola equation: \[ \left(\frac{6x_1}{q}\right)^2 = 36x_1 \] This simplifies to: \[ \frac{36x_1^2}{q^2} = 36x_1 \implies x_1 = \frac{q^2}{x_1} \] Thus, we can solve for \( x_1 \): \[ x_1 = -3p \] ### Step 6: Substitute Back to Find the Locus Substituting \( x_1 \) into the equation: \[ y_1 = \frac{6(-3p)}{q} = -\frac{18p}{q} \] Substituting \( y_1 \) back into the parabola equation: \[ \left(-\frac{18p}{q}\right)^2 = 36(-3p) \implies \frac{324p^2}{q^2} = -108p \] Rearranging gives: \[ 324p^2 + 108pq^2 = 0 \implies q^2 + 3p = 0 \] This leads to the locus equation: \[ y^2 + 3x = 0 \] ### Conclusion The locus of the centroid of the triangle formed by the tangent to the parabola \( y^2 = 36x \) with the coordinate axes is: \[ \boxed{y^2 + 3x = 0} \]

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: Identify the Parabola The given parabola is \( y^2 = 36x \). This can be compared to the standard form \( y^2 = 4ax \), where \( 4a = 36 \). Thus, we find \( a = 9 \). ### Step 2: Write the Equation of the Tangent The equation of a tangent to the parabola \( y^2 = 4ax \) at the point \( (x_1, y_1) \) is given by: \[ ...
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