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

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To find the point of contact between the parabola $$ y^{2} = 4a x $$ and its tangent line $$ y = m x + c $$, follow these steps:

### 1. **Substitute the Tangent into the Parabola**

Start by substituting $$ y = m x + c $$ into the equation of the parabola:

$$
(m x + c)^2 = 4a x
$$

Expanding the left side:

$$
m^{2} x^{2} + 2 m c x + c^{2} = 4a x
$$

Rearranging terms:

$$
m^{2} x^{2} + (2 m c - 4a) x + c^{2} = 0
$$

### 2. **Condition for Tangency**

For the line to be a tangent to the parabola, the quadratic equation in $$ x $$ must have exactly one solution. This occurs when the discriminant is zero:

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

Simplifying the discriminant:

$$
(2m c - 4a)^2 = 4m^{2} c^{2}
$$

Taking the square root of both sides:

$$
2m c - 4a = \pm 2m c
$$

Considering the positive case:

$$
2m c - 4a = 2m c \implies -4a = 0 \quad (\text{Not possible})
$$

Considering the negative case:

$$
2m c - 4a = -2m c \implies 4m c = 4a \implies m c = a
$$

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

Use the parameter $$ t $$ for the parabola $$ y^{2} = 4a x $$, where a general point on the parabola is $$ (a t^{2}, 2a t) $$.

The equation of the tangent at this point is:

$$
y = t x + a t
$$

Comparing this with $$ y = m x + c $$:

$$
m = \frac{1}{t} \quad \text{and} \quad c = a t
$$

Since $$ m c = a $$, substitute $$ t = \frac{1}{m} $$:

$$
c = a \left(\frac{1}{m}\right) = \frac{a}{m}
$$

Now, the coordinates of the point of contact become:

$$
x = a t^{2} = a \left(\frac{1}{m}\right)^{2} = \frac{a}{m^{2}}, \quad y = 2a t = \frac{2a}{m}
$$

### **Final Answer**

The point of contact is $$ \left( \dfrac{a}{m^{2}},\; \dfrac{2a}{m} \right) $$.

$$
\boxed{\left(\dfrac{a}{m^{2}},\; \dfrac{2a}{m}\right)}
$$
The point of contact is $$ \left( \dfrac{a}{m^{2}},\; \dfrac{2a}{m} \right) $$.

The point of contact \( y^{2}=4 a x \) and the tangent \( y=m x+c \) is

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