Question

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?

Answer 1

Coordinates are `(2+sqrt3,-5-4sqrt3)` and `(2-sqrt3,-5+4sqrt3)`

Explanation:

Equation of a line with slope `m` and passing through `(2,0)` is

`y=m(x-2)=mx-2m`

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

`mx-2m=1-x^2` or `x^2+mx-(2m+1)=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+8m+4=0`

wich gives us two values of `m=(-8+-(8^2-4xx1xx4))/2=-4+-2sqrt3`

Hence we have two equations of tangents

`y=-(4+2sqrt3)(x-2)` and

`y=-(4-2sqrt3)(x-2)`

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=-(4+2sqrt3)(x-2)` i.e.

`x^2-(4+2sqrt3)x+7+4sqrt3=0`,

which gives `x=2+sqrt3` and `y=2-(2+sqrt3)^2=-5-4sqrt3`

and `1-x^2=-(4-2sqrt3)(x-2)` i.e.

`x^2-(4-2sqrt3)x+7-4sqrt3=0`,

which gives `x=2-sqrt3` and `y=2-(2-sqrt3)^2=-5+4sqrt3`

Hence coordinates are `(2+sqrt3,-5-4sqrt3)` and `(2-sqrt3,-5+4sqrt3)`