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If \[x = {t^2}\] and \[y = 2t\], then equation of the normal at \[t = 1\] is
A.\[x + y - 3 = 0\]
B.\[x + y - 1 = 0\]
C.\[x + y + 1 = 0\]
D.\[x + y + 3 = 0\]

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Last updated date: 23rd Jul 2024
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Answer
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Hint: Here in this question, we have to find the equation of the normal at the given point \[t = 1\]. Find the equation by using the Point-Slope formula \[y - {y_1} = m\left( {x - {x_1}} \right)\] before finding the equation first we have to find the slope using the formula \[m = - \dfrac{{dx}}{{dy}}\]. On simplification to the point-slope formula we get the required solution.

Complete step by step solution:
Normal form of a straight line, a normal to a line is considered as a line segment drawn from a point that is perpendicular to the given line.
Consider a given two equations:
\[x = {t^2}\]-------(1)
\[y = 2t\]------(2)
Squaring both side in equation (2), we have
\[ \Rightarrow {y^2} = {\left( {2t} \right)^2}\]
\[ \Rightarrow {y^2} = 4{t^2}\]
From equation (1)
\[ \Rightarrow {y^2} = 4x\]------(3)
Now find the point of normal line at \[t = 1\].
Consider equation (1) at \[t = 1\]
\[ \Rightarrow x = {\left( 1 \right)^2}\]
\[\therefore x = 1\]
Substitute x value in equation (3), we have
\[ \Rightarrow {y^2} = 4\left( 1 \right)\]
By taking a square root on both side, we get
\[ \Rightarrow y = \sqrt 4 \]
\[ \Rightarrow y = 2\]
Therefore, the point \[\left( {{x_1},{y_1}} \right) = \left( {1,2} \right)\]
Now, we have to find the equation of the normal to line at the point \[\left( {1,2} \right)\] by using the slope-point formula \[y - {y_1} = m\left( {x - {x_1}} \right)\]-------(4)
Before this, find the slope \[m\]in point-slope formula by using the formula \[m = - \dfrac{{dx}}{{dy}}\] at point \[\left( {1,2} \right)\]
Differentiate equation (3) with respect to x.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {4x} \right)\]
\[ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 4\]
Divide both side by 2y, then
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{2y}}\]
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = \dfrac{4}{{2\left( 2 \right)}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{4}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 1\]
Taking a reciprocal, then
\[ \Rightarrow \dfrac{{dx}}{{dy}} = 1\]
Add – sign on both side, then
\[ \Rightarrow m = - \dfrac{{dx}}{{dy}} = - 1\]
Now we get the gradient or slope of the line which passes through the points \[\left( {1,2} \right)\].
Substitute the slope m and the point \[\left( {{x_1},{y_1}} \right) = \left( {1,2} \right)\] in the point slope formula.
Consider the equation (4)
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Where \[m = - 1\], \[{x_1} = 1\] and \[{y_1} = 2\] on substitution, we get
\[ \Rightarrow y - 2 = - 1\left( {x - 1} \right)\]
\[ \Rightarrow y - 2 = - x + 1\]
Add x on both side, then
\[ \Rightarrow x + y - 2 = 1\]
Subtract 1 on both side, then
\[ \Rightarrow x + y - 2 - 1 = 0\]
On simplification, we get
\[ \Rightarrow x + y - 3 = 0\]
Hence, the equation of the normal of the line is \[x + y - 3 = 0\]. So, the correct answer is “Option A”.

Note: The concept of the equation of normal comes under the concept of application of derivatives. Here the differentiation is applied to the equation of line or curve and then substitute the value of x. We should know about the general equation of a line. Hence these types of problems are solved by the above procedure.