Find the equations to both lines through the point (2,1) that are tangent to the parabola

Question

Find the equations to both lines through the point (2,1) that are tangent to the parabola

y=x²+x+4

 

Answer 1

We need two things to write an equation of any line:
1) a point
2) the slope

We have the point - (2,1) - but we are still missing the slope. We can find the slope by taking the derivative of y=x²+x+4 at the point (2,1), because at the point of tangency, the tangent line and the parabola will have the same slope. 

 

Using basic power derivative rule, the derivative of y=x²+x+4 is 2x+1. Evaluate at x=2 --> 2(2)+1 = 5. Therefore, the slope of our tangent line is 5.

 

Point-slope form for the equation of a line is y-y_o = m(x-x_o), where m is the slope and (x_o, y_o) is a point. So, using m = 5 and (x_o, y_o) = (2, 1), the equation of our tangent line is y-1 = 5(x-2). 

 

We can also rewrite this into slope-intercept form --> y-1 = 5x-10 --> y = 5x-9

 

This is the equation of the line that is tangent to the parabola at (2,1) and therefore passes through the point (2,1). 

 

The alternative (for the second line) is to have a line that is tangent to the parabola at a point other than (2,1) but still passes through the point (2,1). We can find this equation in a similar manner.

 

We will still need the derivative of the parabola: y' = 2x+1

From the equation y=x²+x+4, we can write a general point (x, y), but since y = x²+x+4, we can also write the point as such: (x, x²+x+4)

What we are looking for is the point of tangency. Remember, at the point of tangency, the slope of the tangent line and the slope of the parabola are equal. 

Therefore, for some x, y' will be equal to the slope between the points (x, x²+x+4) and (2,1) [because we want the tangent line to go through both points]

So, we know y', now to find the slope between the two points.

Slope = rise/run = (x²+x+4-1)/(x-2) = (x²+x+3)/(x-2)

Set y' equal to this slope and then solve for x

2x+1 = (x²+x+3)/(x-2)

(2x+1)(x-2) = x²+x+3

2x²-3x-2 = x²+x+3

x²-4x-5=0

(x-5)(x+1)=0

x=5 or x= -1

 

So, the slope of the tangent line is 2(5)+1 = 11 or 2(-1)+1 = -1

Together with the point (2,1), this would generate equations of y-1=11(x-2) or y-1= -1(x-2)