If the tangent to the curve `y= (x)/(x^(2) - 3), x in r ( xne pm sqrt3)`, at a point `(alpha, beta ) ne (0, 0) ` on it is parallel to the line `2x + 6y -11 =0`, then (A) `|6alpha + 2beta|=19` (B) `|6alpha + 2 beta|=9` (C) `|2 alpha + 6 beta |= 19` (D) `| 2 alpha + 6 beta| = 11`

Options

A. `|6alpha + 2beta|=19`
B. `|6alpha + 2 beta|=9`
C. `|2 alpha + 6 beta |= 19`
D. `| 2 alpha + 6 beta| = 11`

Answer

A

Explanation

Equation of given curve is
`y = (x)/(x^(2) - 3), x in R, (x ne pm sqrt3)" " `… (i)
On differentiating Eq. (i) w.r.t x, we get
` (dy)/(dx) = ((x^(2) - 3)- x(2x))/( ( x^(2) - 3)^(2)) = ( (-x^(2) - 3))/((x^(2) - 3)^(2))`
It is given that tangent at a point `( alpha, beta) ne (0, 0)` on it is parallel to the line
`2x + 6y -11 =0`.
` therefore ` Slope of this line = `- (2)/(6)= (dy)/(dx):|_("("alpha, beta")")`
`rArr " " - (alpha ^(2) + 3)/((alpha ^(2)- 3)^(2)) =(1)/(3)`
`rArr 3 alpha ^(2) + 9 = alpha^(4) - 6 alpha ^(2) + 9`
`rArr " " alpha ^(2) - 9 alpha ^(2) = 0`
`rArr " " alpha = 0, -3, 3`
`rArr " " alpha = 3 or -3" " [because alpha ne 0]`
Now, from Eq. (i),
`beta = (alpha )/( alpha ^(2) - 3) rArr beta = (3)/( 9-3) or (-3)/(9-3) = (1)/(2) or - (1)/(2)`
According to the options, `|6alpha + 2 beta| = 19`
at `( alpha, beta)= (pm 3, pm (1)/(2))`