Question

If the straight line \( y = mx + c \) touches the parabola \( y^2 - 4ax + 4a^3 = 0 \), then \( c \) is:

• A. \( am + \frac{a}{m} \)
• B. \( am - \frac{a}{m} \)
• C. \( \frac{a}{m} + a^2m \)
• D. \( \frac{a}{m} - a^2m \)

Hint:

In problems involving tangency, always remember to set the discriminant of the quadratic equation to zero to ensure exactly one solution. This condition guarantees that the line touches the curve.

Answer 1

The Correct Option is D

We are given the parabola equation: \[ y^2 - 4ax + 4a^3 = 0 \] and the straight line equation: \[ y = mx + c \] Since the straight line touches the parabola, the two curves intersect at exactly one point. For this to happen, the discriminant of the quadratic equation formed by substituting \( y = mx + c \) into the parabola equation must be zero. Substitute \( y = mx + c \) into the parabola equation: \[ (mx + c)^2 - 4ax + 4a^3 = 0 \] Expanding the equation: \[ m^2x^2 + 2mcx + c^2 - 4ax + 4a^3 = 0 \] Now, this is a quadratic equation in terms of \( x \): \[ m^2x^2 + (2mc - 4a)x + (c^2 + 4a^3) = 0 \] For this quadratic to have exactly one solution (since the line touches the parabola), the discriminant must be zero: \[ \Delta = (2mc - 4a)^2 - 4m^2(c^2 + 4a^3) = 0 \] Expanding the discriminant: \[ (2mc - 4a)^2 = 4m^2c^2 - 16mac + 16a^2 \] \[ 4m^2(c^2 + 4a^3) = 4m^2c^2 + 16m^2a^3 \] Set the discriminant equal to zero: \[ 4m^2c^2 - 16mac + 16a^2 - 4m^2c^2 - 16m^2a^3 = 0 \] Simplify: \[ -16mac + 16a^2 - 16m^2a^3 = 0 \] Factor out the common terms: \[ -16a \left( mc - a + m^2a^2 \right) = 0 \] Thus, we have: \[ mc - a + m^2a^2 = 0 \] Solve for \( c \): \[ c = a - m^2a^2 \] Therefore, the value of \( c \) is: \[ c = \frac{a}{m} - \frac{a^2}{m} \] Thus, the correct answer is Option D.