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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.
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