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 at the point whose abscissa is 1.
Solution
As per the given information in the question, abscissa means x- axis’s point so x = 1. We can calculate the slope of the curve using dxdy 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,
First of all calculating the coordinates of the point that can be given as ,
Now, calculating the slope of tangent to the given curve at the designated coordinate,
(dxdy)x = 1 = (9x2 - 4)x = 1 = 9(1) - 4 m = 5Hence 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 = 0Above 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=5−1Now, again using point slope method to write the equation of normal,
y - y1 = m(x - x1) ⇒y - 6 = 5−1(x - 1) ⇒5y - 30 = - x + 1 ⇒5y + x = 31 = 0Hence , 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.