Match the column:
Column-IColumn-II(i) Area of the square formed by tangents and normals at the ext…

Question

Match the column:
Column-IColumn-II(i) Area of the square formed by tangents and normals at the extremities of the latus    rectum of the parabola y2 = 4ax(a > 0) is 32, then 'a' is equal to[A] 2(ii)Let f(x) = If Langrange's mean value theorem is applicable in [–2, 2] then m + 3c is equal to[Β] 6(iii) If AFB is a focal chord of the parabola y2 = 4ax and AF = 2, FB = 6 then the length of latus rectum will be (where F is the focus of the parabola) [C] −2(iv) The value of 'a' for which atleast one tangent to the parabola y2 = 4ax becomes normal to the circle x2 + y2 – 2ax – 4ay + 3a2 = 0 is/are[D] 8[Τ] 4

CGP EDU Academic Team · Accepted Answer

The correct answer is (i) [A]; (ii) [E]; (iii) [B]; (iv) [A]. Ans.
(i) [A]
(ii) [E]
(iii) [B]
(iv) [A],[B],[C],[D],[E]
Sol.
(i) Area of the square = 8a 2 = 32 ⇒ a = 2
(ii) f(0 + ) = f(0 – ) ⇒ c = 1, f ′ (0 + ) = f ′ (0 – ) ⇒ m = 1 ∴ m + 3c = 4
(iii) do your self
(iv) ty = x + at 2 is normal to the circle so
2at = a + at 2 ⇒ t
2 – 2t + 1 = 0 ⇒ t = 1 for any value of a so the condition
satisfies ∀ real value of a

Text Solution

Prepare Smarter with CGP Edu

Write a Review

Commentary

Student Reviews

Similar Questions

Column-I	Column-II
(i) Area of the square formed by tangents and normals at the extremities of the latus rectum of the parabola y2 = 4ax(a > 0) is 32, then 'a' is equal to	[A] 2
(ii)Let f(x) = If Langrange's mean value theorem is applicable in [–2, 2] then m + 3c is equal to	[Β] 6
(iii) If AFB is a focal chord of the parabola y2 = 4ax and AF = 2, FB = 6 then the length of latus rectum will be (where F is the focus of the parabola)	[C] −2
(iv) The value of 'a' for which atleast one tangent to the parabola y2 = 4ax becomes normal to the circle x2 + y2 – 2ax – 4ay + 3a2 = 0 is/are	[D] 8
	[Τ] 4

Share this question