Two tangents to the parabola y=1−x2 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+√3,−5−4√3) and (2−√3,−5+4√3)

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−x2, we should have

mx−2m=1−x2 or x2+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 m2−4×1×(−(2m+1)=0

i.e. m2+8m+4=0

wich gives us two values of m=−8±(82−4×1×4)2=−4±2√3

Hence we have two equations of tangents

y=−(4+2√3)(x−2) and

y=−(4−2√3)(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−x2=−(4+2√3)(x−2) i.e.

x2−(4+2√3)x+7+4√3=0,

which gives x=2+√3 and y=2−(2+√3)2=−5−4√3

and 1−x2=−(4−2√3)(x−2) i.e.

x2−(4−2√3)x+7−4√3=0,

which gives x=2−√3 and y=2−(2−√3)2=−5+4√3

Hence coordinates are (2+√3,−5−4√3) and (2−√3,−5+4√3)

Related questions
See all questions in Definition of the Tangent Line
Impact of this question
4901 views around the world
You can reuse this answer
Creative Commons License