Find Equations Of Both Lines Through The Point (2, −3) That Are Tangent To The Parabola Y = X2 + X.

Find the equation of both lines that pass through the point (2, -3) and are tangent to the parabola y = x^2 + x.

Expert Answers
ishpiro eNotes educator| Certified Educator

The slope of the line tangent to the graph of the function `y(x)` at point `(x_0, y_0)` is the value of the derivative of `y(x)` at this point:

`m = y'(x_0)`.

For the given function `y(x) = x^2 + x`, the derivative `y'(x) = 2x + 1`, so the slope of the tangent line is as follows:

`m = 2x_0 + 1`.

If the tangent line passes through the point (2, -3), its equation in point-slope form is as follows:

`y - (-3) = m(x - 2)`.

Since the point `(x_0, y_0)` belongs to both the tangent line and the parabola, then, plugging in `y_0 = x_0^2 + x_0` and `m = 2x_0 + 1`, we get the following:

`x_0^2 + x_0 + 3 = (2x_0 + 1)(x_0 - 2)`

`x_0^2 + x_0 + 3 = 2x_0^2 + x_0 - 4x_0 - 2`

`x_0^2 - 4x_0 - 5 = 0`

`(x_0 - 5)(x_0 + 1) = 0`

From here, `x_0 = 5 `and `x_0 = -1`. So, the slope of the first tangent line is as follows:

`m = 2*5 + 1 = 11`and the slope of the second tangent line is

`m = 2(-1) + 1 = - 1`.

In point-slope form, the equations of the two tangent lines are as follows:

`y + 3 = 11(x - 2)`and `y + 3 = -(x -2)`.

In the slope-intercept form, these are as follows:

`y = 11x - 25` and `y = -x - 1`.

The equations of the lines passing through (2, -3) and tangent to the given parabola are as follows:

y = 11x - 25 and y = - x - 1.

Neethu eNotes educator| Certified Educator

We have to find the equations of both lines through the point (2,-3) that are tangent to the parabola `y=x^2+x`

First we will  find the derivative of y to find the slope m.

i.e. `m=y'=2x+1`

Now we find two points on the two tangent lines to y using: `(x,x^2+x)` and (2,-3). This is to show that (x,f(x)) is a point on the original equation f(x).

Now we have to find the slope of these two points and equate it to the derivative or slope of the original equation.

i.e. `2x+1=\frac{x^2+x+3}{x-2}`
 

implies, `x^2-4x-5=0`
i.e. `x=5 , -1` ``
 
So now at x=5 the slope is m= 2(5)+1=11.So, the equation of the first tangent line through point (2,-3) is y+3=11(x-2)
i.e. `\mathbf{y=11x-25}`
 
at x=-1, the slope is m=2(-1)+1=-1 .So, the equation of the second tangent line through point (2,-3) is: y+3= -1(x-2).
i.e.  `\mathbf{y=-x-1}`

So the equation of the two tangent lines to the parabola `y=x^2+x` passing through the point (2,-3) are:

`y=11x-25` and `y=-x-1`
justaguide eNotes educator| Certified Educator

The equation of the line through (2, -3) and tangential to the parabola y = x^2 + x has to be determined.

The slope of a tangent to a curve y = f(x) at the point x = c is f'(c)

y' = 2x + 1

Let the line that passes through (2, -3) be a tangent to the parabola at (x, x^2 + x)

(x^2 + x + 3)/(x - 2) = 2x + 1

=> x^2 + x + 3 = 2x^2 - 4x + x - 2

=>  x^2 - 4x - 5 = 0

=> x^2 - 5x + x - 5 = 0

=> x(x - 5) + 1(x - 5) = 0

=> (x + 1)(x - 5) = 0

x = -1 and x = 5

The tangents are lines passing through (2, -3) and (-1, 0) and through (2, -3) and (5, 30)

The lines are (y + 3)/(x - 2) = (3/-3) = -1

=> y + 3 = 2 - x

=> x + y + 1 = 0

and (y + 3)/(x  - 2) = (33/3)

=> y + 3 = 11x - 22

=> 11x - y - 25 = 0

The equation of the two tangent lines to the curve y = x^2 + x passing through (2, -3) are x + y + 1 = 0 and 11x - y - 25 = 0