Answer:
Given equation: [tex]y=x^2+1[/tex]
Therefore, we can say that any point on the curve has the coordinates [tex](a, a^2+1)[/tex] (where a is any constant)
To find the gradient of the tangent to the curve at any given point, differentiate the equation.
Given equation:
[tex]y=x^2+1[/tex]
[tex]\implies \dfrac{dy}{dx}=2x[/tex]
Therefore, the gradient at point [tex](a, a^2+1)[/tex] is [tex]2a[/tex]
Using the point-slope form of linear equation, we can create a general equation of the tangent at point [tex](a, a^2+1)[/tex]:
[tex]\begin{aligned}y-y_1 & =m(x-x_1)\\ \implies y-(a^2+1)& =2a(x-a)\end{aligned}[/tex]
[tex]\implies y=2ax-2a^2+a^2+1[/tex]
[tex]\implies y=2ax-a^2+1[/tex]
Given that the tangents pass through point (-1, -3), input this into the general equation of the tangent:
[tex]\begin{aligned}y &=2ax-a^2+1\\ \implies -3 & =2a(-1)-a^2+1\end{aligned}[/tex]
[tex]\implies 0=-2a-a^2+1+3[/tex]
[tex]\implies a^2+2a-4=0[/tex]
Use the quadratic formula to solve for a:
[tex]\implies a=\dfrac{-2\pm\sqrt{2^2-4(1)(-4)}}{2(1)}[/tex]
[tex]\implies a=\dfrac{-2\pm2\sqrt{5}}{2}[/tex]
[tex]\implies a=-1 \pm \sqrt{5}[/tex]
Input the found values of a into the general equation of the tangent to create the equations of the two lines:
[tex]\begin{aligned}a=-1+\sqrt{5}\implies y & =2(-1+\sqrt{5})x-(-1+\sqrt{5})^2+1\\ y & =(-2+2\sqrt{5})x-(6-2\sqrt{5})+1\\ y & =(-2+2\sqrt{5})x+2\sqrt{5}-5 \end{aligned}[/tex]
[tex]\begin{aligned}a=-1-\sqrt{5}\implies y & =2(-1-\sqrt{5})x-(-1-\sqrt{5})^2+1\\ y & =(-2-2\sqrt{5})x-(6+2\sqrt{5})+1\\ y & =(-2-2\sqrt{5})x-2\sqrt{5}-5 \end{aligned}[/tex]
Therefore, the equations of the two lines that pass through the point (-1, -3) and are tangent to the graph of [tex]y=x^2+1[/tex] are:
[tex]y=(-2+2\sqrt{5})x+2\sqrt{5}-5[/tex]
[tex]y=(-2-2\sqrt{5})x-2\sqrt{5}-5[/tex]
