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

Engineering

Mathematics

Tangent and Normal

Question

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

| 6α + 2β | = 9

| 2α + 6β | = 19

| 6α + 2β | = 19

| 2α + 6β | = 11

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${\frac{dy}{dx} |}_{(α, β)} = \frac{- α^{2} - 3}{{(α^{2} - 3)}^{2}}$

Given that :

$\frac{- α^{2} - 3}{{(α^{2} - 3)}^{2}} = - \frac{1}{3}$

⇒ α = 0, ± 3 $(α \neq 0)$

$\Rightarrow β = \pm \frac{1}{2} β \neq 0)$

|6α + 2β| = 19