The point of contact $$ y^{2}=4 a x $$ ar the tangent $$ y=m x+c $$ is

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To find the **point of contact** where the tangent line $$ y = mx + c $$ touches the parabola $$ y^2 = 4ax $$, follow these steps:

### 1. **Condition for Tangency**

For the line to be tangent to the parabola, the system of equations must have exactly one solution. Substitute $$ y = mx + c $$ into the parabola equation:

$$
(mx + c)^2 = 4ax
$$

Expanding and rearranging:

$$
m^2x^2 + 2mcx + c^2 - 4ax = 0
$$

This is a quadratic in $$ x $$. For there to be exactly one solution (tangency condition), the discriminant must be zero:

$$
\Delta = (2mc - 4a)^2 - 4m^2c^2 = 0
$$

Simplifying the discriminant:

$$
4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = -16amc + 16a^2 = 0
$$

$$
-16amc + 16a^2 = 0 \implies mc = a
$$

Thus, the condition is:

$$
c = \frac{a}{m}
$$

### 2. **Finding the Point of Contact**

Now, substitute $$ c = \frac{a}{m} $$ back into the equation of the tangent:

$$
y = mx + \frac{a}{m}
$$

Set this equal to the parabola equation to find the coordinates $$(x, y)$$:

$$
(mx + \frac{a}{m})^2 = 4ax
$$

Expanding and solving for $$ x $$:

$$
m^2x^2 + 2a x + \frac{a^2}{m^2} = 4ax
$$

$$
m^2x^2 - 2ax + \frac{a^2}{m^2} = 0
$$

Using the tangency condition (single solution):

$$
x = \frac{a}{m^2}
$$

Now, find $$ y $$:

$$
y = m \left( \frac{a}{m^2} \right) + \frac{a}{m} = \frac{a}{m} + \frac{a}{m} = \frac{2a}{m}
$$

### **Final Answer**

The point of contact is given by the coordinates:

$$
\left( \frac{a}{m^2},\ \frac{2a}{m} \right)
$$

---

**Therefore, the point of contact where the tangent $$ y = mx + c $$ touches the parabola $$ y^2 = 4ax $$ is $$\left( \frac{a}{m^2}, \frac{2a}{m} \right)$$.**
The point of contact where the tangent $$ y = mx + c $$ touches the parabola $$ y^2 = 4ax $$ is $$ \left( \frac{a}{m^2}, \frac{2a}{m} \right) $$.

The point of contact ar
the tangent is

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