Tangents and Normals: Geometrical Interpretation, Slope, Equation, Orthogonal Curves

Tangents and Normals: Have you ever been on a merry-go-round ride? If yes, you will understand when I say that the force you feel is directed toward the centre of the merry-go-round, yet your velocity (the tendency of motion) is directed in the direction the ride is moving. Another way of expressing the same idea is to say that your velocity is tangential to the ride at all times, but your force is normal to the circle along which you are moving. Can you see a link between the two ways of expressing the same thing? Don’t worry if you don’t; finding tangents and normals to a given curve is what this branch of derivatives application is all about. A tangent is a straight line that just touches the curve at a certain point and does not cross through it. The normal is a straight line that is perpendicular to the tangent.

Geometrical Interpretation of Derivative

Consider the function \(f\left( x \right)\) defined on an open interval \(\left( {a,\,b} \right)\). Let \(P\left( {c,\,f\left( c \right)} \right)\) be a point on the curve \(y = f\left( x \right)\), let us move \(h\) distance away right and left to the point \(P\) then, the coordinates of point \(R\) are \(\left( {c + h,\,\,f\left( {c + h} \right)} \right)\), and the coordinates of point \(Q\) are \(\left( {c – h,\,f\left( {c – h} \right)} \right)\) as shown in the figure given below

Observe that the slope of the chord \(PQ = \frac{{f\left( c \right) – f\left( {c – h} \right)}}{{c – c + h}} = \frac{{f\left( {c – h} \right) – f\left( c \right)}}{{ – h}}\), and the slope of chord \(PR = \frac{{f\left( {c + h} \right) – f\left( c \right)}}{{c + h – c}} = \frac{{f\left( {c + h} \right) – f\left( c \right)}}{c}\)

Now, when the points \(Q\) and \(R\) tends to \(P\),
\(c + h \to c \Rightarrow h \to 0\)
\(c – h \to c \Rightarrow h \to 0\)

Then, we observe that the chord \(PQ\), and \(PR\) becomes tangent to the curve. The slope of the chords \(PQ\) and \(PR\) becomes the slope of the tangent to the curve,
\(y = f\left( x \right)\) at \(P\)

Hence, the slope of the tangent at \(P\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {c + h} \right) – f\left( c \right)}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {c – h} \right) – f\left( c \right)}}{{ – h}}\)

From the definition of differentiability, the slope of the tangent at point \(P\left( {c,f\left( c \right)} \right) = f’\left( c \right) = {\left( {\frac{{dy}}{{dx}}} \right)_P}\)

Slope of Tangents

Let \(y = f\left( x \right)\) be a continuous curve, and let \(P\left( {{x_1},\,{y_1}} \right)\) be a point on it. Then, \({\left( {\frac{{dy}}{{dx}}} \right)_P}\) is the slope of the tangent to the curve\(y = f\left( x \right)\) at \(P\) .

Slope of the tangent at \(P = {\left( {\frac{{dy}}{{dx}}} \right)_P} = \tan \,\psi \), where \(\psi \) is the angle which the tangent at \(P\left( {{x_1},\,{y_1}} \right)\) makes with the positive direction of \(x-\)axis.

Remarks:

• If the tangent at \(P\) is parallel to \(x-\)axis, then
  \(\psi  = 0 \Rightarrow \tan \,\psi  = 0 \Rightarrow\,{\text{Slope}}\,{\text{ = }}\,{\text{0}} \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_P} = 0\)
• If the tangent at \(P\) is perpendicular to \(x-\)axis, or parallel to \(y-\)axis then
  \(\psi  = \frac{\pi }{2} \Rightarrow \tan \,\psi  = {\text{Not}}\,{\text{defined}} \Rightarrow \,{\text{Slope}}\,{\text{ = }}\,{\text{Not}}\,{\text{defined}} \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_P} = {\text{Not}}\,{\text{defined}}\)

Slope of Normals

The normal to a curve at \(P\left( {{x_1},\,{y_1}} \right)\) is a line perpendicular to the tangent at \(P\).

\(\therefore \) Slope of the normal at \(P\) \( =  – \frac{1}{{{\text{Slope}}\,{\text{of}}\,{\text{the}}\,{\text{tanget}}\,{\text{at}}\,P}} =  – \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_P}}} =  – {\left( {\frac{{dx}}{{dy}}} \right)_P}\)

Equation of Tangents and Normals

Equation of a line passing through \(\left( {{x_1},\,{y_1}} \right)\) and having slope \(m\) is
\(y – {y_1} = m\left( {x – {x_1}} \right)\)

Since the slope of the tangent at point \(P\left( {{x_1},\,{y_1}} \right)\) to the curve \(y = f\left( x \right)\) is \({\left( {\frac{{dy}}{{dx}}} \right)_P}\), we get

Equation of the tangent at \(P\left( {{x_1},\,{y_1}} \right)\) to the curve \(y = f\left( x \right)\) is
\(y – {y_1} = {\left( {\frac{{dy}}{{dx}}} \right)_P}\left( {x – {x_1}} \right)\)

Since the normal at \(P\left( {{x_1},\,{y_1}} \right)\) passes through \(P\) and has slope \( – \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_P}}}\). Therefore, the equation of the normal at \(P\left( {{x_1},\,{y_1}} \right)\) to the curve \(y = f\left( x \right)\) is
\(y – {y_1} =  – \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_P}}}\left( {x – {x_1}} \right)\)

Remark \(1\): If \({\left( {\frac{{dy}}{{dx}}} \right)_P} = \infty \), then the tangent at \(\left( {{x_1},\,{y_1}} \right)\) is parallel to \(y-\)axis, and its equation is \(x = {x_1}\).

Remark \(2\):If \({\left( {\frac{{dy}}{{dx}}} \right)_P} = 0\), then the normal at \(\left( {{x_1}\,,{y_1}} \right)\) is parallel to \(y-\)axis,and its equation is \(x = {x_1}\).

Steps to Find the Equation of Tangents and Normals

We may use the following algorithm to find the equations of tangent and normal to a given curve at a given point.

Step 1: Find \(\frac{{dy}}{{dx}}\) from the given equation \(y = f\left( x \right)\).
Step 2: Find the value of \(\frac{{dy}}{{dx}}\) at the given point \(P\left( {{x_1},\,{y_1}} \right)\).
Step 3: If \({\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}}\) is a non-zero finite number, then equation of tangent at \(\left( {{x_1},\,{y_1}} \right)\) is
\(y – {y_1} = {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}}\left( {x – {x_1}} \right)\)
And, equation of normal at \(\left( {{x_1},\,{y_1}} \right)\) is
\(y – {y_1} =  – \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {{x_1},\,{y_1}} \right)}}}}\left( {x – {x_1}} \right)\)
Otherwise, proceed to the next steps.
Step 4: If\({\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1}\,,{y_1}} \right)}} = 0\), then the equations of the tangent and normal at \(\left( {{x_1}\,,{y_1}} \right)\) are \(y = {y_1}\) and \(x = {x_1}\) respectively.
Step 5: If \({\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1}\,,{y_1}} \right)}} =  \pm \infty \), then the equations of the tangent and normal at \(\left( {{x_1}\,,{y_1}} \right)\) are \(x = {x_1}\) and \(y = {y_1}\) respectively.

Angle of Intersection of Two Curves

The angle of intersection of two curves is defined to be the angle between the tangents to the two cures at their point of intersection.

Formula

Let \({C_1}\) and \({C_2}\) be two curves having equations \(y = f\left( x \right)\) and \(y = g\left( x \right)\), respectively. Let \(P{T_1}\) and \(P{T_2}\) be tangents to the curves \({C_1}\) and \({C_2}\), respectively at their point of intersection. Then, the angle \(\phi \) between\(P{T_1}\) and\(P{T_2}\) is the angle of intersection of \({C_1}\) and \({C_2}\).

Let \({\psi _1}\) and \({\psi _2}\) be the angles made by \(P{T_1}\) and \(P{T_2}\) with the positive direction of \(x-\)axis in an anticlockwise direction. Then
\({m_1} = \,\tan {\psi _1}\)
\( \Rightarrow {m_1} = \) slope of tangent to the curve \(y = f\left( x \right)\) at point \(P\)
\(\therefore \,\,{m_1} = \,{\left( {\frac{{dy}}{{dx}}} \right)_P}\)
Similarly, \({m_2} = \,\tan {\psi _2}\)
\( \Rightarrow {m_2} = \) slope of tangent to the curve \(y = g\left( x \right)\) at point \(P\)
\(\therefore \,\,{m_2} = \,{\left( {\frac{{dy}}{{dx}}} \right)_P}\)

From the above figure, we have
\(\phi  + {\psi _2} = {\psi _1}\)
\( \Rightarrow \phi  = {\psi _1} – {\psi _2}\)
\( \Rightarrow \tan \phi  = \,\tan \,\left( {{\psi _1} – {\psi _2}} \right)\)
\( \Rightarrow \tan \phi  = \,\frac{{\tan \,{\psi _1} – \,\tan {\psi _2}}}{{1 + \,\tan \,{\psi _1}\,\tan {\psi _2}}}\)
\(\therefore \tan \phi  = \,\frac{{{m_1} – \,{m_2}}}{{1 + {m_1}\,{m_2}\,}}\)

The alternate angle between the tangents is \({180^ \circ } – \phi \). The angle of intersection is usually regarded to be the lesser of these two angles.

Orthogonal Curves

When two curves intersect at a right angle, they are said to intersect orthogonally, and the curves are referred to as orthogonal curves.

If the curves are orthogonal, then \(\phi  = \frac{\pi }{2}\)
\(\therefore \,\,{m_1}\,{m_2}\, =  – 1\)

Remark: If the angle of intersection of two curves is zero, then \({m_1} = \,{m_2}\) at the point of intersection, and the two curves touch each other at the point of intersection.

Solved Examples – Tangents and Normals

Q.1. Find the equation of the tangent to the curve \(y =  – 5{x^2} + 6x + 7\) at the point \(\left( {\frac{1}{2},\,\frac{{35}}{4}} \right)\).
Solution: The equation of the given curve is
\(y =  – 5{x^2} + 6x + 7\)
\( \Rightarrow \frac{{dy}}{{dx}} =  – 10x + 6\)
\( \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\,\frac{{35}}{4}} \right)}} =  – \frac{{10}}{2} + 6 = 1\)
Thus, the required equation of the tangent at \(\left( {\frac{1}{2},\,\frac{{35}}{4}} \right)\) is
\(y – \frac{{35}}{4} = {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\,\frac{{35}}{4}} \right)}}\left( {x – \frac{1}{2}} \right)\)
\(y – \frac{{35}}{4} = 1\left( {x – \frac{1}{2}} \right)\)
\(\therefore y = x + \frac{{33}}{4}\)

Q.2. Find the equations of the tangent and normal to the parabola \({y^2} = 4ax\) at the point \(\left( {a{t^2},\,2at} \right)\).
Solution: The equation of the given curve is
\({y^2} = 4ax\)       …… \(\left( i \right)\)
Differentiating \(\left( i \right)\) with respect to \(x\), we get
\(2y\frac{{dy}}{{dx}} = 4a\)
\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{2a}}{y}\)
\( \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{t^2},\,2at} \right)}} = \frac{{2a}}{{2at}} = \frac{1}{t}\)
So, the equation of the tangent at \(\left( {a{t^2},\,2at} \right)\) is
\(y – 2at = {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{t^2},\,2at} \right)}}\left( {x – a{t^2}} \right)\)
\( \Rightarrow y – 2at = \frac{1}{t}\left( {x – a{t^2}} \right)\)
\( \Rightarrow ty = x + a{t^2}\)
The equation of the normal at \(\left( {a{t^2},\,2at} \right)\) is
\(y – 2at = \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {a{t^2},\,2at} \right)}}}}\left( {x – a{t^2}} \right)\)
\( \Rightarrow y – 2at =  – \frac{1}{{\frac{1}{t}}}\left( {x – a{t^2}} \right)\)
\( \Rightarrow y – 2at =  – t\left( {x – a{t^2}} \right)\)
\( \Rightarrow y + tx = 2at + a{t^3}\)

Q.3. Find the equation of a tangent line to \(y = 2{x^2} + 7\), that is parallel to the line \(4x – y + 3 = 0\).
Solution: Let the point of contact of the required tangent line be \(\left( {{x_1},\,{y_1}} \right)\)
The equation of the given curve is
\(y = 2{x^2} + 7\)
Differentiating both sides with respect to,\(x\), we get
\(\frac{{dy}}{{dx}} = 4x\)
\(\therefore {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}} = 4{x_1}\)
Since the line \(4x – y + 3 = 0\) is parallel to the tangent at \(\left( {{x_1},\,{y_1}} \right)\).
\(\therefore \) Slope of the tangent at\(\left( {{x_1},\,{y_1}} \right)\) \( = \left( {{\text{Slope}}\,{\text{of}}\,{\text{the}}\,{\text{line}}\,4x – y + 3 = 0} \right)\)
\({\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}} = \frac{{ – 4}}{{ – 1}}\left[ {\because \,{\text{Slope}}\, =  – \frac{{{\text{Coefficient}}\,{\text{of}}\,x}}{{{\text{Coefficient}}\,{\text{of}}\,y}}} \right]\)
\( \Rightarrow 4{x_1} = 4\)
\( \Rightarrow {x_1} = 1\)
Now, \(\left( {{x_1},\,{y_1}} \right)\) lies on \( = 2{x^2} + 7\)
\(\therefore \,{y_1} = 2{x_1^2} + 7 \Rightarrow {y_1} = 2 + 7 = 9\left[ {\because \,{x_1} = 1} \right]\)
So, the coordinates of the point of contact are \(\left( {1,\,9} \right)\).
Hence, the equation of the tangent line is
\(y – 9 = 4\left( {x – 1} \right)\)
\(\therefore \,4x – y + 5 = 0\)

Q.4. Find the points on the curve \(4{x^2} + 9{y^2} = 1\), where the tangents are perpendicular to the line \(2y + x = 0\).
Solution: Let the required point be \(P\left( {{x_1},\,{y_1}} \right)\). The equation of the given curve is
\(4{x^2} + 9{y^2} = 1\)   …….\(\left( i \right)\)
Differentiating both sides with respect to \(x\), we have
\( \Rightarrow 8x + 18y\frac{{dy}}{{dx}} = 0\)
\( \Rightarrow \frac{{dy}}{{dx}} =  – \frac{{4x}}{{9y}}\)
\( \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}} = \frac{{ – 4{x_1}}}{{9{y_1}}}\)
Since tangent at \(\left( {{x_1},\,{y_1}} \right)\) is perpendicular to \(2y + x = 0\). Therefore,
Slope of the tangent at \(\left( {{x_1},\,{y_1}} \right)\) \( \times \) Slope of the line \( =  – 1\)
\( \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}} \times \left( { – \frac{1}{2}} \right) =  – 1\)
\( \Rightarrow \frac{{ – 4{x_1}}}{{9{y_1}}} \times  – \frac{1}{2} =  – 1\)
\(\therefore \,{y_1} = \frac{{ – 2{x_1}}}{9}\)   …..\(\left( ii \right)\)
Since \(P\left( {{x_1},\,{y_1}} \right)\) lies on the curve \((i)\) Therefore,
\(4x_1^2 + 9y_1^2 = 1\)
\( \Rightarrow 4x_1^2 + 9{\left( {\frac{{ – 2{x_1}}}{9}} \right)^2} = 1\)  [Using \((ii)\)]
\( \Rightarrow 4x_1^2 + \frac{{4x_1^2}}{9} = 1 \Rightarrow \,x_1^2 = \frac{9}{{40}} \Rightarrow {x_1} =  \pm \frac{3}{{2\sqrt {10} }}\)
Now, \({x_1} = \frac{3}{{2\sqrt {10} }} \Rightarrow {y_1} = \frac{{ – 2}}{9}\left( {\frac{3}{{2\sqrt {10} }}} \right) =  – \frac{1}{{3\sqrt {10} }}\)   [Using \((ii)\)]
\({x_1} =  – \frac{3}{{2\sqrt {10} }}\)
\( \Rightarrow {y_1} = \frac{{ – 2}}{9}\left( { – \frac{3}{{2\sqrt {10} }}} \right)\)
\(\therefore {y_1} = \frac{1}{{3\sqrt {10} }}\)     [Using \((ii)\)]
Hence, the required points are \(\left( {\frac{3}{{2\sqrt {10} }}.\frac{{ – 1}}{{3\sqrt {10} }}} \right)\) and \(\left( {\frac{{ – 3}}{{2\sqrt {10} }},\,\frac{1}{{3\sqrt {10} }}} \right)\).

Q.5. Find the equations of tangent and normal to the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\), at \(\left( {{x_1},\,{y_1}} \right)\).
Solution: Consider
\(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) ………\((i)\)
Since \(P\left( {{x_1},\,{y_1}} \right)\) lies on the curve, therefore
\(\frac{{{x_1^2}}}{{{a^2}}} + \frac{{{y_1^2}}}{{{b^2}}} = 1\)  ……..\((ii)\)
Differentiating \(i\) with respect to \(x\), we get
\(\frac{{2x}}{{{a^2}}} + \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0 \Rightarrow \frac{{dy}}{{dx}} =  – \frac{{{b^2}x}}{{{a^2}y}} \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}} =  – \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}\)
The equation of the tangent at \(P\left( {{x_1},\,{y_1}} \right)\) is
\(y – {y_1} = {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_1},\,{y_1}} \right)}}\left( {x – {x_1}} \right)\)
\(y – {y_1} =  – \frac{{{b^2}{x_1}}}{{{a^2}{y_1}}}\left( {x – {x_1}} \right)\)
\( \Rightarrow \frac{{y{y_1} – y_1^2}}{{{b^2}}} =  – \left( {\frac{{x{x_1} – x_1^2}}{{{a^2}}}} \right)\)
\( \Rightarrow \frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = \frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}}\)
\( \Rightarrow \frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1\)   [Using \(ii\)]
The equation of the normal at \(P\left( {{x_1},\,{y_1}} \right)\) is
\(y – {y_1} =  – \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {{x_1},\,{y_1}} \right)}}}}\left( {x – {x_1}} \right)\)
\( \Rightarrow y – {y_1} = \frac{{{a^2}{y_1}}}{{{b^2}{x_1}}}\left( {x – {x_1}} \right)\)
\( \Rightarrow \frac{{{b^2}\left( {y – {y_1}} \right)}}{{{y_1}}} = \frac{{{a^2}\left( {x – {x_1}} \right)}}{{{x_1}}}\)
\( \Rightarrow \frac{{{b^2}y}}{{{y_1}}} – {b^2} = \frac{{{a^2}x}}{{{x_1}}} – {a^2}\)
\(\therefore \,\,\frac{{{a^2}x}}{{{x_1}}} – \frac{{{b^2}y}}{{{y_1}}} = {a^2} – {b^2}\)

Summary

The two lines associated with curves are tangents and normals. Each point on the curve can have a tangent, a line that touches the curve at a distinct point. A line perpendicular to the tangent at the point of contact is called a normal. The tangent at point \(\left( {{x_1},\,{y_1}} \right)\) to the curve \(y = f\left( x \right)\) has the equation \(\left( {y – {y_1}} \right) = m\,\left( {x – {x_1}} \right)\), but a normal going through the same point has the equation \(\left( {y – {y_1}} \right) =  – \frac{1}{m}\,\left( {x – {x_1}} \right)\). The slope of the tangent at \(P = {\left( {\frac{{dy}}{{dx}}} \right)_P}\) and that of a normal at \(P = {\left( {\frac{{dx}}{{dy}}} \right)_P}\).The angle between the tangents at the point of intersection of two curves is called the angle of intersection of two curves. When two curves intersect at a right angle, they are called orthogonal curves.

Frequently Asked Questions (FAQs)

Q.1. What is the difference between a tangent line and a normal line?
Ans: A tangent is a straight line that just touches the curve at a certain point and does not cross through it. The normal is a line making an angle of  \({90^ \circ }\) with the tangent at the point of tangency. Also, by the property of perpendicular lines, the slopes of the tangent line and normal line are negative reciprocal of each other.

Q.2. How do you find normals and tangents?
Ans: The equation of the tangent at \(P\left( {{x_1},\,{y_1}} \right)\) to the curve \(y = f\left( x \right)\) is found using the formula,
\( {y – {y_1}} = {\left( {\frac{{dy}}{{dx}}} \right)_P}\,\left( {x – {x_1}} \right)\)
And,
The equation of the normal at \(P\left( {{x_1},\,{y_1}} \right)\) to the curve \(y = f\left( x \right)\) is
\( {y – {y_1}} =  – \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_P}}}\,\left( {x – {x_1}} \right)\)

Q.3. How do you find the equation of normals and tangents?
Ans: The slope of the tangent line to the graph of \(f\) at the particular point \(\left( {{x_0},\,f\left( {{x_0}} \right)} \right)\) on the graph \(y = f\left( x \right)\) is the value \(f’\left( {{x_0}} \right)\) of the derivative \(f’\) of \(f\)
Recall that the equation of  a line with slope \(m\), and point \(\left( {{x_0},\,{y_0}} \right)\) is given by
\(\left( {y – {y_0}} \right) = m\left( {x – {x_0}} \right)\)……….\((i)\)
The slope of the normal to the curve \(y = f\left( x \right)\) at point \(\left( {{x_0},\, {y_0}} \right)\) is \( – \frac{1}{{f’\left( {{x_0}} \right)}}\)
Further, using the formula given in equation \((i)\), we can get the equation of normal.

Q.4. What is a normal in math?
Ans: In geometry, a normal is a line perpendicular to another object, such as a line, ray, or vector. The (infinite) line perpendicular to the tangent line to the curve at the point, for example, is the normal line to a plane curve at a particular point.

Q.5. What is the point of tangency?
Ans: A tangent is a straight line that touches the curve at a certain point and does not cross through it. The point where the tangent and the curve meet is called the point of tangency.

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