Two tangents to a curve intersecting at a point

Question

Q) find the equations of the tangent lines to the curve $y=x^3+x$ which pass through point $(2,2)$

my attempt: I tried to formulating two equations for both tangents and then inserting the values 2,2 in them.After that I tried to solve these equations but there were too many unknowns and little equations

Answer 1

Hint:

If those curves are tangent to the curve, their slope is given by $\;y'=3x^2+1\;$ at each general point $\;(x,\,x^3+x)\;$ on the curve , so for what points $\;(a,\,a^3+a)\;$ on the curve are there lines through them and through $\;(2,2)\;$ whose slope is $\;3a^2+1\;$ ?

Answer 2

HINTS: 1) The slope of the tangent is the value of the derivative of the function at a given point. 2) Any line can be expressed in point-slope form.

Answer 3

Suppose that the tangent at point $(a,a^3+a)$ contains the point $(2,2)$.

We know that the slope of the line is

$$m=3a^2+1$$

but we also know that the slope of the line is

$$ m=\frac{a^3+a-2}{a-2} $$

therefore

$$ \frac{a^3+a-2}{a-2}=3a^2+1 $$

which simplifies to

$$ 2a^2(a-3)=0 $$

So either $a=0$ or $a=3$ giving $m=1$ or $m=28$.

1. \begin{eqnarray} y-2&=&x-2\\ y&=&x \end{eqnarray}

2. \begin{eqnarray} y-2&=&28(x-2)\\ y&=&28x-54 \end{eqnarray}