Question

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

Answer 1

The equations of the tangents that pass through `(2,-3)` are:

`y=-x-1` and
`y = 11x-25`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So for our curve (the parabola) we have

`y=x^2+x`

Differentiating wrt `x` we get:

`dy/dx=2x+1`

Let `P(alpha,beta)` be any generic point on the curve. Then the gradient of the tangent at P is given by:

`m = 2alpha + 1 \ \ \` (using the derivative)

And as P lies on the curve we also have:

`beta = alpha^2+alpha \ \ \` (using the curve equation)

And so the tangent at `P` passes through `(alpha,alpha^2+alpha)` and has gradient `2alpha + 1`, so using the point/slope form `y−y_1=m(x−x_1)` the equation of the tangent at `P` is;

`y - (alpha^2+alpha) = (2alpha+1)(x-alpha)`

if this tangent also passes through `(2,-3)` then;

`\ \-3 - (alpha^2+alpha) = (2alpha+1)(2-alpha)`
`:. -3 - alpha^2-alpha = 3alpha-2alpha^2+2`
`:. alpha^2 -4alpha-5=0`
`:. (alpha-5)(alpha+1)=0`
`:. alpha =-1,5`

If `alpha =-1 => beta = 0`, and the tangent equation becomes:

`y - 0 = (-1)(x+1)`
`:. \ \ y=-x-1`

If `alpha =5 => beta = 30`, and the tangent equation becomes:

`\ \ \ \ \ y - 30 = (11)(x-5)`
`:. y - 30 = 11x-55`
`:. \ \ \ \ \ \ \ \ \y = 11x-25`

Hence the equations of the tangents that pass through `(2,-3)` are
`y=-x-1` and `y = 11x-25`

We can confirm this graphically:

Answer 2

The two tgts. are `11x-y-25=0, and, x+y+1=0`.

Explanation:

Let `m` be the slope of tgt. line thro. the pt. `(2,-3).`

Using the Slope-point Form , we get, its eqn., `y+3=m(x-2).`

Solving this eqn. with `y=x^2+x`, we get the Pts. of Intersection of

the Parabola with the tgt. line.

`x^2+x=y=m(x-2)-3=mx-2m-3, i.e.,`

`x^2+x(1-m)+(2m+3)=0..............(star).`

`(star)` being a quadratic eqn. in `x`, it can have at most `2` roots that may give the `x-`co-ords. of the Pts. of Int.

But, a tgt. essentially intersects the Parabola in only one pt., `(star)` must have `2` equal roots, meaning that, `Delta=0," for "(star).`

`rArr (1-m)^2=4(1)(2m+3)`

`rArr m^2-10m-11=0 rArr m=11, or, m=-1`.

For `m=11," the tgt. is "y+3=11(x-2)=11x-22, i.e., 11x-y-25=0,` and, If

`m=-1," the tgt. is, "x+y+1=0;` as Respected Steve has obtained!

N.B.:
`(1): m=-1,&, (star) rArr x^2+2x+1=0 rArr x=-1, so, y=x^2+x=0.`
`:." the pt. of contact is "(-1,1).`

`(2): m=11" gives "(5,30)" as the pt. of contact."`

Enjoy Maths.!