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Q.Find the equations of the tangent and normal to the parabola y2=8xy^2 = 8x at t=12t = \dfrac{1}{2}.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 6mImportance★★★★★
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Identify aa from y2=8xy^2=8x, locate the point for t=12t=\tfrac12, then apply the standard tangent and normal formulas for a parabola in parametric form.

  1. Compare y2=8xy^2=8x with y2=4axy^2=4ax: 4a=8⇒a=24a=8\Rightarrow a=2.
  2. Parametric coordinates on the parabola: (x,y)=(at2,2at)=(2t2,4t)(x,y)=(at^2,2at)=(2t^2,4t).
  3. At t=12t=\tfrac12: x=2(12)2=12x=2\left(\tfrac12\right)^2=\tfrac12, y=4(12)=2y=4\left(\tfrac12\right)=2. So the point is (12,2)\left(\tfrac12,2\right).
  4. Standard tangent to y2=4axy^2=4ax at parameter tt: ty=x+at2ty=x+at^2.
  5. With a=2,t=12a=2,t=\tfrac12: 12y=x+2(14)=x+12\tfrac12y=x+2\left(\tfrac14\right)=x+\tfrac12. Multiply by 2: y=2x+1y=2x+1, i.e. 2x−y+1=02x-y+1=0.
  6. Standard normal to y2=4axy^2=4ax at parameter tt: y+tx=2at+at3y+tx=2at+at^3. …

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