JEE Main 2019 · 10 April · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

If the tangent to the curve y=xx2−3y = \frac{x}{x^2 - 3}, x∈Rx \in R, (x≠±3)(x \neq \pm\sqrt{3}), at a point (α,β)≠(0,0)(\alpha, \beta) \neq (0, 0) on it is parallel to the line 2x+6y−11=02x + 6y - 11 = 0, then
  1. A.
    ∣6α+2β∣=19|6\alpha + 2\beta| = 19
  2. B.
    ∣6α+2β∣=9|6\alpha + 2\beta| = 9
  3. C.
    ∣2α+6β∣=19|2\alpha + 6\beta| = 19
  4. D.
    ∣2α+6β∣=11|2\alpha + 6\beta| = 11
Show answer

Correct option: A

[JEE Main 2019 · 10 April · Shift II]

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