Question:
The normal drawn to the parabola
\[
y^2=8x
\]
at \((2,4)\) meets the parabola again at the point
The normal drawn to the parabola
\[
y^2=8x
\]
at \((2,4)\) meets the parabola again at the point
Show Hint
For the parabola
\[
y^2=4ax,
\]
the normal at parameter \(t\) is
\[
y=-tx+2at+at^3.
\]
After obtaining the normal, substitute it into the parabola and use the known point to identify the second point of intersection.
Updated On: Jul 29, 2026
- \((18,12)\)
- \((18,-12)\)
- \((40.5,18)\)
- \((40.5,-18)\)
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The Correct Option is B
Solution and Explanation
Concept:
For the parabola
\[
y^2=4ax,
\]
a point on the parabola can be written as
\[
(at^2,\,2at).
\]
The equation of the normal at parameter \(t\) is
\[
y=-tx+2at+at^3.
\]
Step 1: Find the parameter corresponding to \((2,4)\). Given \[ y^2=8x. \] Comparing with \[ y^2=4ax, \] we get \[ a=2. \] The point \((2,4)\) lies on the parabola. Using \[ (at^2,2at), \] we have \[ 2t^2=2. \] \[ t^2=1. \] Since \[ 2at=4, \] \[ 4t=4. \] \[ t=1. \]
Step 2: Find the equation of the normal at \(t=1\). Using \[ y=-tx+2at+at^3, \] with \[ a=2,\qquad t=1, \] \[ y=-x+4+2. \] \[ y=-x+6. \] Thus the normal is \[ x+y-6=0. \]
Step 3: Find the other point of intersection with the parabola. Substitute \[ y=6-x \] into \[ y^2=8x. \] \[ (6-x)^2=8x. \] \[ x^2-12x+36=8x. \] \[ x^2-20x+36=0. \] \[ (x-2)(x-18)=0. \] Thus, \[ x=2 \] or \[ x=18. \] The point \(x=2\) corresponds to the given point \((2,4)\). Hence the second point is obtained from \[ y=6-18=-12. \] Therefore, \[ (18,-12). \]
Step 4: Write the final answer. \[ \boxed{(18,-12)} \]
Step 1: Find the parameter corresponding to \((2,4)\). Given \[ y^2=8x. \] Comparing with \[ y^2=4ax, \] we get \[ a=2. \] The point \((2,4)\) lies on the parabola. Using \[ (at^2,2at), \] we have \[ 2t^2=2. \] \[ t^2=1. \] Since \[ 2at=4, \] \[ 4t=4. \] \[ t=1. \]
Step 2: Find the equation of the normal at \(t=1\). Using \[ y=-tx+2at+at^3, \] with \[ a=2,\qquad t=1, \] \[ y=-x+4+2. \] \[ y=-x+6. \] Thus the normal is \[ x+y-6=0. \]
Step 3: Find the other point of intersection with the parabola. Substitute \[ y=6-x \] into \[ y^2=8x. \] \[ (6-x)^2=8x. \] \[ x^2-12x+36=8x. \] \[ x^2-20x+36=0. \] \[ (x-2)(x-18)=0. \] Thus, \[ x=2 \] or \[ x=18. \] The point \(x=2\) corresponds to the given point \((2,4)\). Hence the second point is obtained from \[ y=6-18=-12. \] Therefore, \[ (18,-12). \]
Step 4: Write the final answer. \[ \boxed{(18,-12)} \]
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