Question

If the straight line \[ lx+my+n=0 \] touches the parabola \[ y^2=4ax, \] then

• A. \(al^2=mn\)
• B. \(am^2=ln\)
• C. \(an^2=ml\)
• D. \(a^2m=l^2n\)

Hint:

For the parabola \[ y^2=4ax, \] the tangent in slope form is \[ y=mx+\frac{a}{m}. \] This is the quickest way to solve tangent-condition MCQs.

Answer 1

The Correct Option is B

Step 1: Use the condition for a line and parabola to meet once.
The line $lx+my+n=0$ touches $y^2=4ax$ when they have exactly one common point, so substitution should give a perfect-square (repeated-root) quadratic.

Step 2: Express $x$ from the line.
From $lx+my+n=0$ we get $x=-\dfrac{my+n}{l}$.

Step 3: Substitute into the parabola.
Put this in $y^2=4ax$: $y^2=4a\left(-\dfrac{my+n}{l}\right)$, which gives $ly^2+4amy+4an=0$.

Step 4: Apply the tangency (equal roots) condition.
For one repeated root the discriminant is zero: $(4am)^2-4\cdot l\cdot 4an=0$.

Step 5: Simplify.
$16a^2m^2-16aln=0$. Divide by $16a$: $am^2-ln=0$.

Step 6: State the relation.
Hence $am^2=ln$. \[ \boxed{am^2=ln} \]