Question

The condition that the line \(lx+my=1\) may be normal to the curve \(y^2=4ax\), is

• A. \(al^3-2alm^2=m^2\)
• B. \(al^2+2alm^3=m^2\)
• C. \(al^3+2alm^2=m^3\)
• D. \(al^3+2alm^2=m^2\)

Hint:

Normal to \(y^2=4ax\): \(y=-tx+2at+at^3\). Match slope and intercept with \(lx+my=1\) to get the condition.

Answer 1

The Correct Option is D

Step 1: Parametric form of parabola.
For \(y^2=4ax\), parametric point is:
\[ (at^2,2at) \] Step 2: Equation of normal in parametric form.
Normal to parabola at parameter \(t\) is:
\[ y=-tx+2at+at^3 \] Step 3: Convert given line to slope form.
Given line:
\[ lx+my=1 \Rightarrow y=-\frac{l}{m}x+\frac{1}{m} \] So slope of line is \(-\frac{l}{m}\).
Step 4: Match slope with normal slope.
Normal slope is \(-t\).
So:
\[ -t=-\frac{l}{m}\Rightarrow t=\frac{l}{m} \] Step 5: Compare intercepts.
Normal equation:
\[ y=-tx+2at+at^3 \Rightarrow \text{intercept}=2at+at^3 \] Given line intercept = \(\frac{1}{m}\).
So:
\[ 2at+at^3=\frac{1}{m} \] Substitute \(t=\frac{l}{m}\):
\[ 2a\frac{l}{m}+a\left(\frac{l}{m}\right)^3=\frac{1}{m} \] Multiply by \(m^3\):
\[ 2alm^2+al^3=m^2 \] \[ al^3+2alm^2=m^2 \] Final Answer: \[ \boxed{al^3+2alm^2=m^2} \]