if tangent to curve `2y^(3)=ax^(2)+x^(3)` at point (a,a) cuts off intercepts `alpha, beta` on co-ordinate axes where `alpha^(2)+ beta^(2)=61` then the value of 'a' is equal to

Options

A. `pm 30`
B. `pm 5`
C. `pm 6`
D. `pm 61`

Answer

A

Explanation

We have,
`2y^(3) =ax^(2) =x^(3)`
`rArr 6y^(2) (dy)/(dx)=2ax + 3x^(2) rArr (dy)/(dx) = (2ax+3x^(2))/(6y^(2)) rArr ((dy)/(dx))_((a","a))=(5)/(6)`
The equation of the tangent at (a, a) is
`y-a=(5)/(6)(x-a)rArr 5x-6y +a=0`
This intercepts lenths `-a//5 and a//6` with a and y-axis respectively.
`therefore alpha = -a//5 and Beta = a//6`
Now,
`alpha^(2)+beta^(2)=61 rArr (a^(2))/(25)+(a^(2))/(36)=61 rArr a^(2)=25 xx 36 rArr a=pm 30`