Question

Find the point on the curve \[y = {x^3} - 11x + 5\]at which the tangent is\[y = x - 11\].

Answer 1

Hint:At any point on the curve the slope of the tangent is equal to the slope of the curve. In this question from the given equation of the curve and the equation of the tangent we will find their respective slope and as we know at any point on the curve the slope of the tangent is equal to the slope of the curve so by equating their slope we will find the point on which the line is tangent to the curve.

Complete step by step solution:
Given the equation of the curve \[y = {x^3} - 11x + 5\]
The tangent \[y = x - 11\]

Now let us assume the point of contact is\[\left( {x,y} \right)\]
Now as we know at any point on the curve the slope of the tangent is equal to the slope of the curve so we will find the slope of respective curves,

For the given equation of the tangent\[y = x - 11\], we know the general equation of a straight line is given as\[y = mx + c\], hence by comparing the two equations we can say the slope of the tangent is\[{m_{\tan }} = 1\].

Now we know \[\dfrac{{dy}}{{dx}}\]of an equation gives the slope of a curve, hence we differentiate the given equation of the curve to find its slope as
\[m = \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^3} - 11x + 5} \right)}}{{dx}}\]
By differentiating the equation we get
\[
{m_{curve}} = \dfrac{{d\left( {{x^3}} \right)}}{{dx}} - \dfrac{{d\left( {11x} \right)}}{{dx}} +
\dfrac{{d\left( 5 \right)}}{{dx}} \\
\Rightarrow {m_{curve}} = 3{x^2} - 11 \\
\]

Now since we know the slope of the tangent is equal to the slope of the curve at any point on curve so we can write
\[
{m_{curve}} = {m_{\tan }} \\
\Rightarrow 3{x^2} - 11 = 1 \\

\Rightarrow 3{x^2} = 12 \\
\Rightarrow {x^2} = 4 \\
\Rightarrow x = \pm 2 \\
\]

Now substitute the value of \[x\] in the curve equation to find the point of intersection
When\[x = + 2\], \[y = {x^3} - 11x + 5 = {\left( 2 \right)^3} - 11 \times 2 + 5 = - 9\]

When\[x = - 2\], \[y = {x^3} - 11x + 5 = {\left( { - 2} \right)^3} - 11 \times \left( { - 2} \right) + 5 = 19\]
So the required points are \[\left( {2, - 9} \right)\]and\[\left( { - 2,19} \right)\].

But we can see the point \[\left( { - 2,19} \right)\]does not satisfy the equation\[y = x - 11\],

hence we can say \[\left( {2, - 9} \right)\]is the point on the curve \[y = {x^3} - 11x + 5\]at which the tangent\[y = x - 11\].

Note:Since a tangent drawn on a curve is a straight line so we use the general form of equation of a straight line to find the equation of the tangent given in the form\[y = mx + c\], here in the equation of the line \[m\]tells about the slope of the line or the slope of the curve which is intersecting a curve.