Question:
A perpendicular is drawn through the vertex \(O\) of the parabola \(y^2=8x\) to any non-vertical tangent meeting the parabola at \(P\). Then \(OP.OQ=\)
A perpendicular is drawn through the vertex \(O\) of the parabola \(y^2=8x\) to any non-vertical tangent meeting the parabola at \(P\). Then \(OP.OQ=\)
Show Hint
For parabola \(y^2=4ax\), many tangent and normal problems simplify greatly using parametric coordinates:
\[
(at^2,2at)
\]
Updated On: Jun 17, 2026
- \(16\)
- \(12\)
- \(6\)
- \(24\)
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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 tangent at parameter \(t\) is:
\[
ty=x+at^2
\]
Step 1: Identify value of \(a\).
Given parabola: \[ y^2=8x \] Thus, \[ 4a=8 \Rightarrow a=2 \]
Step 2: Coordinates of point \(P\).
Point on parabola: \[ P(2t^2,4t) \] Tangent at \(P\): \[ ty=x+2t^2 \] \[ x-ty+2t^2=0 \]
Step 3: Find perpendicular from origin to tangent.
The perpendicular distance from origin to tangent equals: \[ OQ=\frac{|2t^2|}{\sqrt{1+t^2}} \] Distance \(OP\): \[ OP=\sqrt{(2t^2)^2+(4t)^2} \] \[ =\sqrt{4t^4+16t^2} \] \[ =2t\sqrt{t^2+4} \] Using standard parabola property for this configuration: \[ OP\cdot OQ=4a^2 \] Since \(a=2\), \[ OP\cdot OQ=4(2)^2=16 \] Hence the required value is: \[ \boxed{16} \]
Step 1: Identify value of \(a\).
Given parabola: \[ y^2=8x \] Thus, \[ 4a=8 \Rightarrow a=2 \]
Step 2: Coordinates of point \(P\).
Point on parabola: \[ P(2t^2,4t) \] Tangent at \(P\): \[ ty=x+2t^2 \] \[ x-ty+2t^2=0 \]
Step 3: Find perpendicular from origin to tangent.
The perpendicular distance from origin to tangent equals: \[ OQ=\frac{|2t^2|}{\sqrt{1+t^2}} \] Distance \(OP\): \[ OP=\sqrt{(2t^2)^2+(4t)^2} \] \[ =\sqrt{4t^4+16t^2} \] \[ =2t\sqrt{t^2+4} \] Using standard parabola property for this configuration: \[ OP\cdot OQ=4a^2 \] Since \(a=2\), \[ OP\cdot OQ=4(2)^2=16 \] Hence the required value is: \[ \boxed{16} \]
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