Question

Find thentangent lines to the curve y^2-x+1=0 at the points (2,-1) and (2,1)?

Answer 1

tangent line at `(2, -1): " "y = -1/2x`
tangent line at `(2, 1): " "y = 1/2x`

Explanation:

Given: `y^2-x+1=0` Find tangent lines at `(2, -1), " & "(2,1)`

The derivative is a slope function.
Find the first derivative using implicit differentiation:

`2yy' -1 = 0`

`2yy' = 1`

`y' = 1/(2y)`

Use: `y - y_1 = m (x - x_1)`

Find the tangent line at `(2, -1):`

`m_1 = y'(2, -1) = 1/(2(-1)) = -1/2`

`y - (-1) = -1/2 (x - 2)`

`y + 1 = -1/2 x + 1 " "=> " "y = -1/2x`

Tangent line at `(2, 1):`

`m_2 = y'(2, 1) = 1/(2(1)) = 1/2`

`y - 1 = 1/2 (x - 2)`

`y - 1 = 1/2x -1" "=> " "y = 1/2x`

Answer 2

Tangents are `x+2y=0` and `x-2y=0`

Explanation:

Slope of a tangent line to a curve `f(x,y)=0` at a point on the curve is given by the value of `(dy)/(dx)` at that point . We can find `(dy)/(dx)` using implicit differentiation, but before that we must check whether given points lie on curve `f(x.y)` or not.

Here point `(2,-1)` and `(2,1)` satisfy the equation `y^2-x+1=0` and hence they lie on the curve `y^2-x+1=0`.

Now differentiating `y^2-x+1=0`, we get

`2y(dy)/(dx)-1=0` or `(dy)/(dx)=1/(2y)`

Hence, tangent at `(2,-1)` has slope `1/(2*(-1))=-1/2`

and tangent is `y-(-1)=-1/2(x-2)` or `x+2y=0`

and tangent at `(2,1)` has slope `1/(2*1)=1/2`

and tangent is `y-1=1/2(x-2)` or `x-2y=0`

graph{(y^2-x+1)(x-2y)(x+2y)=0 [-1.434, 8.566, -2.32, 2.68]}