Let C be a curve given by `y=1 + sqrt(4x-3), x gt (3)/(4)` . If P is a point on C such that the tangent at P has slope `(2)/(3)`, then a point through which the normal at P passes, is

Options

A. `(3, -4)`
B. `(1, 7)`
C. `(4, -3)`
D. `(2,3)`

Answer

B

Explanation

Let `P(alpha, 1+sqrt(4alpha -3)` be a point on C, where `alpha gt (3)/(4)` such that tangent at P has slope `(2)/(3)` .
Now, `y=1+sqrt(4x-3)`
`rArr (dy)/(dx) = (2)/(sqrt(4x-3)) rArr ((dy)/(dx))_(p)=(2)/(sqrt(4alpha -3))`
It is given that `((dy)/(dx))_(p)=(2)/(3)`
`rArr (2)/(sqrt(4alpha -3)) = (2)/(3) rArr 4 alpha -3 =9 rArr alpha =3`
So, the coordinates of P are (3, 4).
The equation of the normal at P(3, 4) is
`y-4= -(3)/(2)(x-3) or, 3x+2y-17=0`
Clearly, it basses through (1, 7).