Q.The line $lx + my + n = 0$ will touch the parabola $y^2 = 4ax$ if $ln = am^2$.

Concept: Condition for a line to be tangent to a parabola.

A line touches a parabola when the two equations have exactly one common solution — i.e. the resulting quadratic has discriminant zero.

From the line $lx + my + n = 0$, write $x = -\dfrac{my+n}{l}$ and substitute into $y^2 = 4ax$:

For tangency, its discriminant must vanish: …

The claim is that the line $lx + my + n = 0$ is tangent to the parabola $y^2 = 4ax$ exactly when $ln = am^2$. "Touches" means tangent — the line and the curve share exactly one common point. Algebraically, if we solve the two equations together we get a quadratic, and "exactly one solution" means its discriminant is zero. Let us derive the condition and check it against the statement.

Step 1 — Set up the intersection

We want the points common to the line and the parabola. From the line $lx + my + n = 0$, express $x$ in terms of $y$ (taking $l \neq 0$, the case of a genuine slanted/vertical tangent):

Step 2 — Substitute into the parabola

Put this $x$ into $y^2 = 4ax$:

Multiply through by $l$ and collect all terms on one side:

This is a quadratic in $y$, of the form $Ay^2 + By + C = 0$ with

Step 3 — Impose tangency (discriminant $= 0$)

The line touches the parabola when this quadratic has a repeated root, i.e. its discriminant vanishes:

Expand:

Step 4 — Simplify

Factor out $16a$:

$16a\left(am^2 - ln\right) = 0.$ …