Tangent lines from a point not on the curve

When the tangent starts from a point off the curve, the point of tangency is unknown, so first write it as $(t,\ f(t))$. The tangent there is

It passes through $(p,\ q)$, so substitute $x=p$, $y=q$ and solve $q-f(t)=f'(t)(p-t)$ for $t$.

For example, for $y=x^2$ and the point $(0,\ -4)$, the tangent is $y=2t(x-t)+t^2$, and $-4=-2t^2+t^2$ gives $t=\pm 2$. The two tangents are $y=4x-4$ and $y=-4x-4$.

Examples

1. 01What is the slope of the tangent at this point of tangency?$\displaystyle y=x^{2}, \quad (t,\ t^{2})$AnswerHide

   Answer$\displaystyle 2t$

   Steps

   1. $\displaystyle y$$\displaystyle = x^{2}$
   2. Differentiate
   3. $\displaystyle y'$$\displaystyle = 2x$
   4. Its value at x = t is the slope
   5. $\displaystyle y'(t)$$\displaystyle = 2t$
2. 02Which equation states that this tangent passes through (0, -2)?$\displaystyle y = 2t(x-t)+t^{2}$AnswerHide

   Answer$\displaystyle -2 = -2t^{2}+t^{2}$

   Steps

   1. $\displaystyle 2t(0-t)+t^{2}$$\displaystyle = -2t^{2}+t^{2}$
   2. Passing through means this equals -2
3. 03Solve for t.$\displaystyle -t^{2} = -16$AnswerHide

   Answer$\displaystyle t=\pm 4$

   Steps

   1. $\displaystyle -t^{2} = -16$$\displaystyle \Rightarrow t^{2} = 16$
   2. Two numbers square to 16
   3. $\displaystyle t$$\displaystyle = \pm 4$
   4. That is why there are two tangents
4. 04Write the point of tangency on this curve as (t, □). What goes in □?$\displaystyle y=x^{2}$AnswerHide

   Answer$\displaystyle t^{2}$

   Steps

   1. It is on the curve, so put t into the equation
   2. $\displaystyle y$$\displaystyle = x^{2}$$\displaystyle x = t$$\displaystyle \Rightarrow y = t^{2}$
5. 05Which gives the slope of the tangent at $x=-2$?$\displaystyle f(x)=x^{2}$A$\displaystyle -2$B$\displaystyle f'(x)$C$\displaystyle f(-2)$D$\displaystyle f'(-2)$AnswerHide

   Answer$\displaystyle f'(-2)$
6. 06Tangent line at $x=2$. What goes in □?$\displaystyle f(x)=x^{2} \;\Rightarrow\; y=\square(x-2)+4$AnswerHide

   Answer$\displaystyle 4$

FAQ

Q1Can I put the given point $(p,\ q)$ straight into the tangent formula?AnswerHide

A

No. In $y-f(a)=f'(a)(x-a)$, the point $(a,\ f(a))$ is the point of tangency on the curve. A point off the curve is not a point of tangency, so set the point of tangency as $t$ first and then use the condition that the line passes through $(p,\ q)$.

Q2How many tangents can be drawn?AnswerHide

A

There is one point of tangency for each real solution of the equation in $t$. For $y=x^2$, different values of $t$ give different tangents, so two solutions mean two tangents.