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

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