# The feet of the normals to y^2 = 4ax from the point (6a, 0) are: (0, 0) (4a, -4a) (4a, -4a) (0, 0), (4a, 4a), (4a, -4a)

## Question

The feet of the normals to $y^2 = 4ax$ from the point $(6a, 0)$ are:

$(0, 0)$

$(4a, -4a)$

$(4a, -4a)$

$(0, 0), (4a, 4a), (4a, -4a)$

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## Answer

To find the feet of the normals to the parabola given by ( y^2 = 4ax ) from the point ( (6a, 0) ), we need to follow a systematic approach involving the properties of the parabola and the equations of the normals.

**Steps to Find the Feet of the Normals:**

**Equation of the Parabola and its Derivation**:The given parabola is ( y^2 = 4ax ).

The parametric form for this parabola is ( (at^2, 2at) ).

**Normal to the Parabola**:For the point ((at^2, 2at)), the equation of the normal is: $$ y + tx = 2at + at^3 $$ Plugging in the coordinates of the given point ((6a, 0)), we get: $$ 0 + t \cdot 6a = 2at + at^3 $$

Simplifying, we obtain: $$ 6at = 2at + at^3 \quad \Rightarrow \quad 4at = at^3 \quad \Rightarrow \quad t(t^2 - 4) = 0 $$

**Solving for ( t )**:The equation ( t(t^2 - 4) = 0 ) gives us three solutions: ( t = 0 ), ( t = 2 ), and ( t = -2 ).

**Finding the Feet of the Normals**:For ( t = 0 ): [ (at^2, 2at) = (0, 0) ]

For ( t = 2 ): [ (at^2, 2at) = (4a, 4a) ]

For ( t = -2 ): [ (at^2, 2at) = (4a, -4a) ]

The **feet of the normals** to the parabola ( y^2 = 4ax ) from the point ((6a, 0)) are ( (0, 0) ), ( (4a, 4a) ), and ( (4a, -4a) ).

Thus, the correct answer is:
**(0, 0), (4a, 4a), (4a, -4a)**.

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