Question:
The equation of the normal to the parabola
\[
y^2=12x
\]
at the point \((3\lambda^2,6\lambda)\) is
The equation of the normal to the parabola
\[
y^2=12x
\]
at the point \((3\lambda^2,6\lambda)\) is
Show Hint
For the parabola \(y^2=4ax\), remember:
\[
\text{Tangent: } ty=x+at^2,
\]
\[
\text{Normal: } y=-tx+2at+at^3.
\]
These formulas save considerable time in examinations.
Updated On: Jun 10, 2026
- \[ y=-\lambda x+6\lambda+3\lambda^3 \]
- \[ y=-\lambda x+6\lambda+3\lambda^2 \]
- \[ y=-2\lambda x+6\lambda+3\lambda^3 \]
- \[ y=-\lambda x+3\lambda+6\lambda^3 \]
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For the parabola
\[
y^2=4ax,
\]
the parametric coordinates of a point are
\[
(at^2,2at).
\]
The equation of the normal at parameter \(t\) is
\[
y=-tx+2at+at^3.
\]
This standard result is frequently used in coordinate geometry.
Step 1: Identify the parabola Given \[ y^2=12x. \] Comparing with \[ y^2=4ax, \] we obtain \[ 4a=12. \] Therefore, \[ a=3. \]
Step 2: Identify the parameter The given point is \[ (3\lambda^2,6\lambda). \] For \(a=3\), \[ (at^2,2at) = (3t^2,6t). \] Comparing, \[ t=\lambda. \]
Step 3: Use the normal formula Substituting \[ a=3, \qquad t=\lambda, \] into \[ y=-tx+2at+at^3, \] we get \[ y=-\lambda x+6\lambda+3\lambda^3. \] Hence, \[ \boxed{y=-\lambda x+6\lambda+3\lambda^3}. \] Therefore the correct option is \[ \boxed{(A)}. \]
Step 1: Identify the parabola Given \[ y^2=12x. \] Comparing with \[ y^2=4ax, \] we obtain \[ 4a=12. \] Therefore, \[ a=3. \]
Step 2: Identify the parameter The given point is \[ (3\lambda^2,6\lambda). \] For \(a=3\), \[ (at^2,2at) = (3t^2,6t). \] Comparing, \[ t=\lambda. \]
Step 3: Use the normal formula Substituting \[ a=3, \qquad t=\lambda, \] into \[ y=-tx+2at+at^3, \] we get \[ y=-\lambda x+6\lambda+3\lambda^3. \] Hence, \[ \boxed{y=-\lambda x+6\lambda+3\lambda^3}. \] Therefore the correct option is \[ \boxed{(A)}. \]
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