If the tangent to the curve y = (x/x2 - 3) , x ∈ R , (x ≠ ± √3), at a point (α , β) ≠ (0, 0) on it is parallel to the line 2x + 6y - 11 = 0, then :

Question

If the tangent to the curve y = (x/x2 - 3) , x ∈ R , (x ≠ ± √3), at a point (α , β) ≠ (0, 0) on it is parallel to the line 2x + 6y - 11 = 0, then : (A) |6 α + 2 β | = 19 (B) |2 α + 6 β | = 11  (C) |6 α + 2 β | = 9 (D) |2 α + 6 β | = 19

Tardigrade Admin · Accepted Answer

(dy/dx) Bigg|(α , β) = (-α2 -3/(α2-3)2) Given that: (-α2 -3/(α2 - 3)2) = (1/3) ⇒ α =0 ±3 (α ≠0 ) ⇒ β=± (1/2) . (β≠0) |6α+2β|=19

Q. If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, (x \neq = \pm 3)$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

$∣6 α + 2 β ∣ = 19$

$∣2 α + 6 β ∣ = 11$

$∣6 α + 2 β ∣ = 9$

$∣2 α + 6 β ∣ = 19$

$\frac{d y}{d x} ∣ ∣_{(α, β)} = \frac{- α ^{2} - 3}{( α ^{2} - 3 ) ^{2}}$
Given that :
$\frac{- α ^{2} - 3}{( α ^{2} - 3 ) ^{2}} = \frac{1}{3}$
$\Rightarrow α = 0 \pm 3 (α \neq = 0)$
$\Rightarrow β = \pm \frac{1}{2} . (β \neq = 0)$
$∣ 6 α + 2 β ∣ = 19$

Q. If the tangent to the curve y=x2−3x​,x∈R,(x=±3​), at a point (α,β)=(0,0) on it is parallel to the line 2x+6y−11=0, then :

∣6α+2β∣=19

∣2α+6β∣=11

∣6α+2β∣=9

∣2α+6β∣=19