Question

Determine the equations of two lines that pass through the point (-1,-3) and are tangent

to the graph of y=x² +1.

Answer 1

Given equation:
`y=x^2+1`

Therefore, we can say that any point on the curve has the coordinates
`(a, a^2+1)` (where a is any constant)

To find the gradient of the tangent to the curve at any given point, differentiate the equation.

Given equation:

`y=x^2+1`

`\implies (dy)/(dx)=2x`

Therefore, the gradient at point
`(a, a^2+1)` is
`2a`

Using the point-slope form of linear equation, we can create a general equation of the tangent at point
`(a, a^2+1)`:

`\begin{aligned}y-y_1 & =m(x-x_1)\\ \implies y-(a^2+1)& =2a(x-a)\end{aligned}`

`\implies y=2ax-2a^2+a^2+1`

`\implies y=2ax-a^2+1`

Given that the tangents pass through point (-1, -3), input this into the general equation of the tangent:

`\begin{aligned}y &=2ax-a^2+1\\ \implies -3 & =2a(-1)-a^2+1\end{aligned}`

`\implies 0=-2a-a^2+1+3`

`\implies a^2+2a-4=0`

Use the quadratic formula to solve for a:

`\implies a=(-2\pm√(2^2-4(1)(-4)))/(2(1))`

`\implies a=(-2\pm2√(5))/(2)`

`\implies a=-1 \pm √(5)`

Input the found values of a into the general equation of the tangent to create the equations of the two lines:

`\begin{aligned}a=-1+√(5)\implies y & =2(-1+√(5))x-(-1+√(5))^2+1\\ y & =(-2+2√(5))x-(6-2√(5))+1\\ y & =(-2+2√(5))x+2√(5)-5 \end{aligned}`

`\begin{aligned}a=-1-√(5)\implies y & =2(-1-√(5))x-(-1-√(5))^2+1\\ y & =(-2-2√(5))x-(6+2√(5))+1\\ y & =(-2-2√(5))x-2√(5)-5 \end{aligned}`

Therefore, the equations of the two lines that pass through the point (-1, -3) and are tangent to the graph of
`y=x^2+1` are:

`y=(-2+2√(5))x+2√(5)-5`

`y=(-2-2√(5))x-2√(5)-5`