# What are the equations of the tangent lines out P(2,1) to the parabola with equation y = x^2? Thank you!

Jun 11, 2018

Two tangents

• $y = 2 \left(2 - \sqrt{3}\right) x - 7 + 4 \sqrt{3}$
• $y = 2 \left(2 + \sqrt{3}\right) x - 7 - 4 \sqrt{3}$

#### Explanation:

The tangent to a curve $y \left(x\right)$ at the point $\left({x}_{0} , {y}_{0}\right)$ is given by

$y - {y}_{0} = | \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = {x}_{0}} \left(x - {x}_{0}\right)$

For $y = {x}^{2}$ this becomes

$y - {y}_{0} = 2 {x}_{0} \left(x - {x}_{0}\right)$

with ${y}_{0} = {x}_{0}^{2}$, so that

$y = 2 {x}_{0} x - {x}_{0}^{2}$

For the tangent to go through $\left(2 , 1\right)$ we must have

$1 = 4 {x}_{0} - x - {x}_{0}^{2} \implies$
${x}_{0}^{2} - 4 {x}_{0} + 1 = 0 \implies$
${\left({x}_{0} - 2\right)}^{2} = 3 \implies$
${x}_{0} = 2 \pm \sqrt{3}$

So, there are two tangents to the parabola from $\left(2 , 1\right)$ to the parabola $y = {x}^{2}$ and they are

• for ${x}_{0} = 2 - \sqrt{3}$, $y = 2 \left(2 - \sqrt{3}\right) x - 7 + 4 \sqrt{3}$
• for ${x}_{0} = 2 + \sqrt{3}$, $y = 2 \left(2 + \sqrt{3}\right) x - 7 - 4 \sqrt{3}$