Let x = 2t, y = t²/3 be a conic. Let S be the focus and B be the point on the axis of the conic such that SA ⊥ BA, where A is any point on the conic. If k is the ordinate of the centroid of the ΔSAB, then lim _(t → 1) k is equal to :

Solution

Maths sir · 4 steps

Turn the parametric pair into x² = 12y, find B from a dot product, and then put t = 1.

1. Identify the conic. Put t = x/2 into y = t²/3 to get x² = 12y. This parabola has 4a = 12, so a = 3. Its axis is the y-axis and its focus is S(0, 3).
2. Condition for B. Take A(2t, t²/3) and B(0, b) on the axis. SA ⊥ BA means the dot product of SA⃗ = (2t, t²/3 - 3) and BA⃗ = (2t, t²/3 - b) is zero.

   4t² + (t²/3 - 3)(t²/3 - b) = 0 ⇒ b = t²/3 + 12t²/(t² - 9)

3. Centroid ordinate. k = 1/3(3 + t²/3 + b). This expression is continuous at t = 1, so the limit is its value at t = 1. There, b = 1/3 - 12/8 = -7/6.
4. So. k = 1/3(3 + 1/3 - 7/6) = 1/3 × 13/6 = 13/18, which is option (d).

The idea to keep. Whenever a conic is given parametrically, eliminate the parameter first to read off the focus and axis. For a perpendicularity condition, set the dot product of the two vectors to zero.

Answer from the NTA final answer key. Solution by PrepWiser.