Solveeit Logo

Question

Question: Find the equation of tangent and normal to the curve \[{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 4x...

Find the equation of tangent and normal to the curve 3x3 - 4x + 7{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}} at the point whose abscissa is 11.

Explanation

Solution

As per the given information in the question, abscissa means x- axis’s point so x  = 1{\text{ = 1}}. We can calculate the slope of the curve using dydx\dfrac{{{\text{dy}}}}{{{\text{dx}}}} method. And also remember the information that tangent and normal are perpendicular to each other. And finally, we need to write the equation of the line using point-slope form.

Complete step by step solution: Given curve 3x3 - 4x + 7{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}},
First of all calculating the coordinates of the point that can be given as ,

x = 1, y = 3x3 - 4x + 7  = 3 - 4 + 7  = 6 (x,y) = (1,6)  {\text{x = 1,}} \\\ {\text{y = 3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}} \\\ {\text{ = 3 - 4 + 7}} \\\ {\text{ = 6}} \\\ {\text{(x,y) = (1,6)}} \\\

Now, calculating the slope of tangent to the given curve at the designated coordinate,

(dydx)x = 1 = (9x2 - 4)x = 1  = 9(1) - 4 m = 5  {{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}_{{\text{x = 1}}}}{\text{ = (9}}{{\text{x}}^{\text{2}}}{\text{ - 4}}{{\text{)}}_{{\text{x = 1}}}} \\\ {\text{ = 9(1) - 4}} \\\ {\text{m = 5}} \\\

Hence as we know the slope and point so we can write the equation of line as ,

y - y1 = m(x - x1) y - 6 = 5(x - 1) y - 6 = 5x - 5 5x - y + 1 = 0  {\text{y - }}{{\text{y}}_{\text{1}}}{\text{ = m(x - }}{{\text{x}}_{\text{1}}}{\text{)}} \\\ \Rightarrow {\text{y - 6 = 5(x - 1)}} \\\ \Rightarrow {\text{y - 6 = 5x - 5}} \\\ \Rightarrow {\text{5x - y + 1 = 0}} \\\

Above is the equation of tangent and now using the perpendicular condition to calculate the slope of normal’s line.

m1m2=1 as, m1=5 m2=15  {{\text{m}}_1}{m_2} = - 1 \\\ {\text{as, }}{m_1} = 5 \\\ \Rightarrow {m_2} = \dfrac{{ - 1}}{5} \\\

Now, again using point slope method to write the equation of normal,

y - y1 = m(x - x1) y - 6 = 15(x - 1) 5y - 30 = - x + 1 5y + x = 31 = 0  {\text{y - }}{{\text{y}}_{\text{1}}}{\text{ = m(x - }}{{\text{x}}_{\text{1}}}{\text{)}} \\\ \Rightarrow {\text{y - 6 = }}\dfrac{{ - 1}}{5}{\text{(x - 1)}} \\\ \Rightarrow {\text{5y - 30 = - x + 1}} \\\ \Rightarrow {\text{5y + x = 31 = 0}} \\\

Hence , 5y - 30 = - x + 1{\text{5y - 30 = - x + 1}} is equation of normal.

Note: In common usage, the abscissa refers to the horizontal (x) axis and the ordinate refers to the vertical (y) axis of a standard two-dimensional graph.
A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point. A normal to a curve is a line perpendicular to a tangent to the curve.