Q.If the line $y = mx + 1$ is tangent to the parabola $y^2 = 4x$ then find the value of $m$.

The key idea here is the tangency condition for a line and a parabola.

1. The given parabola is $y^2 = 4x$. Comparing this to the standard form $y^2 = 4ax$, we identify $4a = 4$, which means $a = 1$.
2. The given line is $y = mx + 1$. Comparing this to the standard form $y = mx + c$, we identify $c = 1$. …

When a line is tangent to a curve, it means the line touches the curve at exactly one point. This geometric idea has a direct algebraic translation: if we substitute the equation of the line into the equation of the curve, we should get a quadratic equation with exactly one solution. This happens when the discriminant of the quadratic equation is zero.

For standard conic sections like parabolas, there are specific conditions derived from this principle that directly relate the parameters of the line and the curve. These conditions are very useful for quickly solving problems involving tangency.

Let's use the standard tangency condition for a parabola.

1. Identify the given equations:

   We are given the equation of a line:

   $y = mx + 1 \quad \text{(Equation 1)}$

   And the equation of a parabola:

   $y^2 = 4x \quad \text{(Equation 2)}$

2. Recall the standard form of the parabola and its tangency condition:

   The standard form of a parabola opening to the right, with its vertex at the origin, is $y^2 = 4ax$.

   The condition for a line $y = mx + c$ to be tangent to the parabola $y^2 = 4ax$ is given by:

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

   ›Proof

   Derivation of the Tangency Condition

   To understand why this condition holds, we can substitute the line equation into the parabola equation.

   1. Substitute $y = mx + c$ into $y^2 = 4ax$: $(mx + c)^2 = 4ax$
   2. Expand and rearrange into a quadratic equation in $x$: $m^2x^2 + 2mcx + c^2 = 4ax$ $m^2x^2 + (2mc - 4a)x + c^2 = 0$
   3. For the line to be tangent, there must be exactly one point of intersection. This means the quadratic equation must have exactly one real root, which occurs when its discriminant is zero. The discriminant $\Delta$ of a quadratic $Ax^2 + Bx + C = 0$ is $B^2 - 4AC$. Here, $A = m^2$, $B = (2mc - 4a)$, and $C = c^2$. So, $\Delta = (2mc - 4a)^2 - 4(m^2)(c^2) = 0$
   4. Expand and simplify: $4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = 0$ $-16amc + 16a^2 = 0$
   5. Divide by $16a$ (assuming $a \neq 0$, which is true for a parabola): $-mc + a = 0$ $a = mc$
   6. Solve for $c$: $c = \frac{a}{m}$ This is the required tangency condition. Note that $m \neq 0$ for this condition to be valid. If $m=0$, the line is horizontal ($y=c$), and a horizontal line can only be tangent to $y^2=4ax$ if $c=0$, meaning $y=0$ (the x-axis), which is tangent at the vertex. However, $y=0$ is not of the form $y=mx+1$.

3. Determine the value of 'a' for the given parabola: …