Question

How do you find the equation of both lines through the point (2, -3) that are tangent to the parabola y=x^2-x? And then show that there is no tangent line through the point (2, 7)?

Answer 1

Any tangent line to
`y=x^2-x` at a point
`(a,b)=(a,a^2-a)` will have slope
`y'(a)=2x-1\bigg_(x=a)=2a-1`. Such a line has equation


`y-(a^2-a)=(2a-1)(x-a)\implies y=(2a-1)x-a^2`

For this (these) line(s) to pass through (2, -3), you must have


`-3=(2a-1)*2-a^2\implies a^2-4a-1=0\implies a=2\pm\sqrt5`

So the two lines in questions are
`y=(2a-1)x-a^2` for these values of
`a`.

To show there is no tangent line that passes through (2, 7), you have to show there is no solution for
`a` in


`7=(2a-1)*2-a^2`

You have


`a^2-4a+9=0`

but this has no real solutions, so no such tangent line exists.