Skip to content
Question 88 of 126

Q.The point of intersection of the tangents at t1=tt_1 = t and t2=3tt_2 = 3t to the parabola y2=8xy^2 = 8x is :

(a) (t2,4t)(t^2, 4t)
(b) (6t2,8t)(6t^2, 8t)
(c) (4t,t2)(4t, t^2)
(d) (8t,6t2)(8t, 6t^2)
Puducherry TnboardTamil Nadu HSC (DGE) Board 2018MCQ· 1mImportance★★★★★
70% · 88/126 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Using the standard result that tangents to y2=4axy^2=4ax at parameters t1,t2t_1,t_2 meet at (at1t2, a(t1+t2))(at_1t_2,\,a(t_1+t_2)), with a=2a=2, t1=tt_1=t, t2=3tt_2=3t gives the intersection point (6t2,8t)(6t^2,8t).

  1. Compare y2=8xy^2=8x with the standard form y2=4axy^2=4ax: 4a=8⇒a=24a=8\Rightarrow a=2.
  2. A point on the parabola with parameter tit_i is (ati2,2ati)=(2ti2,4ti)(at_i^2,2at_i) = (2t_i^2,4t_i), and the tangent there is tiy=x+ati2=x+2ti2t_iy = x + at_i^2 = x + 2t_i^2.
  3. Tangent at t1=tt_1=t: ty=x+2t2ty = x + 2t^2 ... (I)
  4. Tangent at t2=3tt_2=3t: 3ty=x+2(3t)2=x+18t23ty = x + 2(3t)^2 = x+18t^2 ... (II)
  5. Subtract (I) from (II): 2ty=16t2⇒y=8t2ty = 16t^2 \Rightarrow y = 8t (for t≠0t\ne0). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.