# Two tangents to the parabola y=1-x^2 intersect at point (2,0). Find the coordinates of points on parabola on which these are tangents?

Mar 15, 2017

Coordinates are $\left(2 + \sqrt{3} , - 5 - 4 \sqrt{3}\right)$ and $\left(2 - \sqrt{3} , - 5 + 4 \sqrt{3}\right)$

#### Explanation:

Equation of a line with slope $m$ and passing through $\left(2 , 0\right)$ is

$y = m \left(x - 2\right) = m x - 2 m$

As this line will normally have two common points with the parabola $y = 1 - {x}^{2}$, we should have

$m x - 2 m = 1 - {x}^{2}$ or ${x}^{2} + m x - \left(2 m + 1\right) = 0$

The two roots of this equation will give two points, but for a tangent, we need two coincident points i.e. roots should be equal, which is possible only when determinant is zero.

Hence, for tangent, we should have m^2-4xx1xx(-(2m+1)=0

i.e. ${m}^{2} + 8 m + 4 = 0$

wich gives us two values of $m = \frac{- 8 \pm \left({8}^{2} - 4 \times 1 \times 4\right)}{2} = - 4 \pm 2 \sqrt{3}$

Hence we have two equations of tangents

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

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

graph{(y+(4+2sqrt3)(x-2))(y+(4-2sqrt3)(x-2))(y-1+x^2)=0 [-8.585, 11.415, -8.4, 1.6]}

Coordinates will be given by $1 - {x}^{2} = - \left(4 + 2 \sqrt{3}\right) \left(x - 2\right)$ i.e.

${x}^{2} - \left(4 + 2 \sqrt{3}\right) x + 7 + 4 \sqrt{3} = 0$,

which gives $x = 2 + \sqrt{3}$ and $y = 2 - {\left(2 + \sqrt{3}\right)}^{2} = - 5 - 4 \sqrt{3}$

and $1 - {x}^{2} = - \left(4 - 2 \sqrt{3}\right) \left(x - 2\right)$ i.e.

${x}^{2} - \left(4 - 2 \sqrt{3}\right) x + 7 - 4 \sqrt{3} = 0$,

which gives $x = 2 - \sqrt{3}$ and $y = 2 - {\left(2 - \sqrt{3}\right)}^{2} = - 5 + 4 \sqrt{3}$

Hence coordinates are $\left(2 + \sqrt{3} , - 5 - 4 \sqrt{3}\right)$ and $\left(2 - \sqrt{3} , - 5 + 4 \sqrt{3}\right)$