Find the tangent and the normal to the curve y^2 – 6x^2 + 4y + 19 = 0 at the point (2, 1).

To find the tangent and normal to the curve y^2 – 6x^2 +
4y + 19 = 0 at the point (2, 1).

We know that the tangent
and normal to the curve at (x1,y1) is given by. y-y1 = m(x-x1)..(1) and y-y1 =
(-1/m)(x-x1)...(2), where m is the dy/dx at
(x1,y1).

(x1,y1) = (2,1)..

We
differentiate y^2 – 6x^2 + 4y + 19 = 0  to find the value of dy/dx at the point (2,
1).

=> 2ydy/dx -12x+4dy/dx =
0.

= (2y+4)dy/dx = 12x.

dy/dx
= m = (12x)/(2y+4) = 12*2/(2*1+1) = 8, as  (x,y) =
(2,1).

Therefore m = 8. Substituting in the equation of the
tangent and normal at (1) and (2), we get:

Tangent : y-1 =
8(x-2). Or y = 8x-16+1 .

Or y = 8x
-15.

Or 8x - y - 15 =
0.

Normal: y-1
(-1/8)(x-2).

8y - 1 =
-(x-2),

x+8y -3 =
0.