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

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### 2 Answers

First we have to find dy/dx for the curve.

dy/dx(y^2 – 6x^2 + 4y + 19) = 0

=> 2y dy/dx – 12x + 4dy/dx = 0

=> y dy/dx – 6x + 2dy/dx = 0

=> dy/dx = 6x /(y+2)

Now at the point (2, 1) dy/dx = 12/3 = 4. This is equal to the slope of the tangent at this point.

y – 1 = 4(x – 2)

The slope of the normal at this point is the inverse reciprocal or -1/4

y – 1 = (-1/4) (x – 2)

**Therefore the required tangent is y – 1 = 4(x – 2) and the required normal is y – 1 = (-1/4) (x – 2).**

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.**