Find the equation of tangent and normal to the curve \[{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}}\] at the point whose abscissa is \[1\].
Answer
623.1k+ views
Hint: As per the given information in the question, abscissa means x- axis’s point so x \[{\text{ = 1}}\]. We can calculate the slope of the curve using \[\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\] 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 \[{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}}\],
First of all calculating the coordinates of the point that can be given as ,
\[
{\text{x = 1,}} \\
{\text{y = 3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}} \\
{\text{ = 3 - 4 + 7}} \\
{\text{ = 6}} \\
{\text{(x,y) = (1,6)}} \\
\]
Now, calculating the slope of tangent to the given curve at the designated coordinate,
\[
{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}_{{\text{x = 1}}}}{\text{ = (9}}{{\text{x}}^{\text{2}}}{\text{ - 4}}{{\text{)}}_{{\text{x = 1}}}} \\
{\text{ = 9(1) - 4}} \\
{\text{m = 5}} \\
\]
Hence as we know the slope and point so we can write the equation of line as ,
\[
{\text{y - }}{{\text{y}}_{\text{1}}}{\text{ = m(x - }}{{\text{x}}_{\text{1}}}{\text{)}} \\
\Rightarrow {\text{y - 6 = 5(x - 1)}} \\
\Rightarrow {\text{y - 6 = 5x - 5}} \\
\Rightarrow {\text{5x - y + 1 = 0}} \\
\]
Above is the equation of tangent and now using the perpendicular condition to calculate the slope of normal’s line.
\[
{{\text{m}}_1}{m_2} = - 1 \\
{\text{as, }}{m_1} = 5 \\
\Rightarrow {m_2} = \dfrac{{ - 1}}{5} \\
\]
Now, again using point slope method to write the equation of normal,
\[
{\text{y - }}{{\text{y}}_{\text{1}}}{\text{ = m(x - }}{{\text{x}}_{\text{1}}}{\text{)}} \\
\Rightarrow {\text{y - 6 = }}\dfrac{{ - 1}}{5}{\text{(x - 1)}} \\
\Rightarrow {\text{5y - 30 = - x + 1}} \\
\Rightarrow {\text{5y + x = 31 = 0}} \\
\]
Hence , \[{\text{5y - 30 = - x + 1}}\] 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.
Complete step by step solution: Given curve \[{\text{3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}}\],
First of all calculating the coordinates of the point that can be given as ,
\[
{\text{x = 1,}} \\
{\text{y = 3}}{{\text{x}}^{\text{3}}}{\text{ - 4x + 7}} \\
{\text{ = 3 - 4 + 7}} \\
{\text{ = 6}} \\
{\text{(x,y) = (1,6)}} \\
\]
Now, calculating the slope of tangent to the given curve at the designated coordinate,
\[
{{\text{(}}\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{)}}_{{\text{x = 1}}}}{\text{ = (9}}{{\text{x}}^{\text{2}}}{\text{ - 4}}{{\text{)}}_{{\text{x = 1}}}} \\
{\text{ = 9(1) - 4}} \\
{\text{m = 5}} \\
\]
Hence as we know the slope and point so we can write the equation of line as ,
\[
{\text{y - }}{{\text{y}}_{\text{1}}}{\text{ = m(x - }}{{\text{x}}_{\text{1}}}{\text{)}} \\
\Rightarrow {\text{y - 6 = 5(x - 1)}} \\
\Rightarrow {\text{y - 6 = 5x - 5}} \\
\Rightarrow {\text{5x - y + 1 = 0}} \\
\]
Above is the equation of tangent and now using the perpendicular condition to calculate the slope of normal’s line.
\[
{{\text{m}}_1}{m_2} = - 1 \\
{\text{as, }}{m_1} = 5 \\
\Rightarrow {m_2} = \dfrac{{ - 1}}{5} \\
\]
Now, again using point slope method to write the equation of normal,
\[
{\text{y - }}{{\text{y}}_{\text{1}}}{\text{ = m(x - }}{{\text{x}}_{\text{1}}}{\text{)}} \\
\Rightarrow {\text{y - 6 = }}\dfrac{{ - 1}}{5}{\text{(x - 1)}} \\
\Rightarrow {\text{5y - 30 = - x + 1}} \\
\Rightarrow {\text{5y + x = 31 = 0}} \\
\]
Hence , \[{\text{5y - 30 = - x + 1}}\] 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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

A dentist uses a small mirror that gives a magnification class 12 physics CBSE

