Question

the tangent to the curve 𝑦 = 𝑥² − 5𝑥 − 2 at the point (1,-6) intersects the normal to

the same curve at the point (2, -8) at point P . find the coordinates of P

Answer 1

Solution

Given that

$y = {x^2} - 5x - 2$

The gradient of the tangent is

${m_t} = \frac{{dy}}{{dx}} = 2x - 5$

The gradient of the tangent at the point (1, - 6) is

${m_t} = 2\left( 1 \right) - 5 = - 3$

Hence equation of the tangent at the point (1, -6) is

$\begin{gathered} y - {y_1} = {m_t}\left( {x - {x_1}} \right) \\ y - \left( { - 6} \right) = - 3\left( {x - 1} \right) \\ y + 6 = - 3x + 3 \\ \end{gathered}$

$3x + y + 3 = 0$ .... (1)

Now we know the gradient of the tangent is ${m_t} = \frac{{dy}}{{dx}} = 2x - 5$

Therefore, the gradient of the normal will be ${m_n} = - \frac{1}{{{m_t}}} = - \frac{1}{{2x - 5}}$

And gradient of the normal at the point (2, -8) is ${m_n} = - \frac{1}{{2\left( 2 \right) - 5}} = - \frac{1}{{ - 1}} = 1$

Therefore, the equation of the normal at the point (2, -8) is 

$\begin{gathered} y - {y_1} = {m_n}\left( {x - {x_1}} \right) \\ y - \left( { - 8} \right) = 1\left( {x - 2} \right) \\ y + 8 = x - 2 \\ \end{gathered}$

$x - y - 10 = 0$ .... (2)

Now the solving (1) and (2), we get 

$\begin{gathered} 3x + y + 3 + x - y - 10 = 0\\ 4x - 7 = 0 \\ x = \frac{7}{4} \\ \end{gathered}$

Replacing in (2)

$\begin{gathered} \frac{7}{4} - y - 10 = 0 & \\ y = - \frac{{33}}{4} \\ \end{gathered}$  

Hence the point of intersection of the tangent and the normal is

 

$\left( {x,y} \right):\left( {\tfrac{7}{4},\,\,\tfrac{{ - 33}}{4}} \right) = \left( {1.75,\, - 8.25} \right)$

The tangent, normal, and the coordinates of the point P, where the tangent and the normal intersect are shown in the plot below.