Question:
If the straight line
\[
lx+my+n=0
\]
touches the parabola
\[
y^2=4ax,
\]
then
If the straight line
\[
lx+my+n=0
\]
touches the parabola
\[
y^2=4ax,
\]
then
Show 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.
Updated On: Jun 16, 2026
- \(al^2=mn\)
- \(am^2=ln\)
- \(an^2=ml\)
- \(a^2m=l^2n\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The tangent to the parabola
\[
y^2=4ax
\]
having slope \(m_1\) is
\[
y=m_1x+\frac{a}{m_1}.
\]
A line will be tangent to the parabola if its equation can be reduced to this form.
Step 1: Write the given line in slope form. \[ lx+my+n=0 \] \[ y=-\frac{l}{m}x-\frac{n}{m} \] Hence slope \[ m_1=-\frac{l}{m}. \]
Step 2: Compare with the tangent form. For a tangent, \[ \text{Intercept} = \frac{a}{m_1} \] Therefore, \[ -\frac{n}{m} = \frac{a}{-\frac{l}{m}} = -\frac{am}{l} \] \[ \frac{n}{m} = \frac{am}{l} \] \[ ln=am^2 \] \[\begin{aligned} \boxed{am^2=ln} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
Step 1: Write the given line in slope form. \[ lx+my+n=0 \] \[ y=-\frac{l}{m}x-\frac{n}{m} \] Hence slope \[ m_1=-\frac{l}{m}. \]
Step 2: Compare with the tangent form. For a tangent, \[ \text{Intercept} = \frac{a}{m_1} \] Therefore, \[ -\frac{n}{m} = \frac{a}{-\frac{l}{m}} = -\frac{am}{l} \] \[ \frac{n}{m} = \frac{am}{l} \] \[ ln=am^2 \] \[\begin{aligned} \boxed{am^2=ln} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
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