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

Feb 6, 2017

The equations of the tangents that pass through $\left(2 , - 3\right)$ are:

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

#### 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:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 1$

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

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

And as P lies on the curve we also have:

$\beta = {\alpha}^{2} + \alpha \setminus \setminus \setminus$ (using the curve equation)

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

$y - \left({\alpha}^{2} + \alpha\right) = \left(2 \alpha + 1\right) \left(x - \alpha\right)$

if this tangent also passes through $\left(2 , - 3\right)$ then;

$\setminus \setminus - 3 - \left({\alpha}^{2} + \alpha\right) = \left(2 \alpha + 1\right) \left(2 - \alpha\right)$
$\therefore - 3 - {\alpha}^{2} - \alpha = 3 \alpha - 2 {\alpha}^{2} + 2$
$\therefore {\alpha}^{2} - 4 \alpha - 5 = 0$
$\therefore \left(\alpha - 5\right) \left(\alpha + 1\right) = 0$
$\therefore \alpha = - 1 , 5$

If $\alpha = - 1 \implies \beta = 0$, and the tangent equation becomes:

$y - 0 = \left(- 1\right) \left(x + 1\right)$
$\therefore \setminus \setminus y = - x - 1$

If $\alpha = 5 \implies \beta = 30$, and the tangent equation becomes:

$\setminus \setminus \setminus \setminus \setminus y - 30 = \left(11\right) \left(x - 5\right)$
$\therefore y - 30 = 11 x - 55$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus y = 11 x - 25$

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

We can confirm this graphically:

Feb 6, 2017

The two tgts. are $11 x - y - 25 = 0 , \mathmr{and} , x + y + 1 = 0$.

#### Explanation:

Let $m$ be the slope of tgt. line thro. the pt. $\left(2 , - 3\right) .$

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

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

the Parabola with the tgt. line.

${x}^{2} + x = y = m \left(x - 2\right) - 3 = m x - 2 m - 3 , i . e . ,$

${x}^{2} + x \left(1 - m\right) + \left(2 m + 3\right) = 0. \ldots \ldots \ldots \ldots . \left(\star\right) .$

$\left(\star\right)$ 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., $\left(\star\right)$ must have $2$ equal roots, meaning that, $\Delta = 0 , \text{ for } \left(\star\right) .$

$\Rightarrow {\left(1 - m\right)}^{2} = 4 \left(1\right) \left(2 m + 3\right)$

$\Rightarrow {m}^{2} - 10 m - 11 = 0 \Rightarrow m = 11 , \mathmr{and} , m = - 1$.

For $m = 11 , \text{ the tgt. is } y + 3 = 11 \left(x - 2\right) = 11 x - 22 , i . e . , 11 x - y - 25 = 0 ,$ 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.
$\therefore \text{ the pt. of contact is } \left(- 1 , 1\right) .$

$\left(2\right) : m = 11 \text{ gives "(5,30)" as the pt. of contact.}$

Enjoy Maths.!