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?

1 Answer
Shwetank Mauria
Mar 15, 2017

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)#

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