Question

Find the equation of the tangent line to the curve \(y = x^2 − 2x + 7\) which is:
(a) parallel to the line \(2x − y + 9 = 0 \)
(b) perpendicular to the line \(5y − 15x = 13\)

Answer 1

The equation of the given curve is y=x²-2x+7
On differentiating with respect to x, we get:
\(\frac {dy}{dx}\) = 2x-2

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(a) The equation of the line is 2x − y + 9 = 0.

2x − y + 9 = 0 
y = 2x + 9 
This is of the form y = mx + c
∴ Slope of the line = 2 If a tangent is parallel to the line 2x − y + 9 = 0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have:
2 = 2x − 2
2x=4
x=2
Now, x = 2
y = 4 − 4 + 7 = 7 
Thus, the equation of the tangent passing through (2, 7) is given by,
y-7=2(x-2)
y-2x-3=0

Hence, the equation of the tangent line to the given curve (which is parallel to line 2x − y + 9 = 0) is y-2x-3=0 .

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(b) The equation of the line is 5y − 15x = 13

5y-15x=13 
y=3x+\(\frac {13}{5}\)
5y − 15x = 13
∴ This is of the form y = mx + c. 
∴Slope of the line = 3 If a tangent is perpendicular to the line 5y − 15x = 13, then the slope of the tangent is
\(-\frac {1}{slop\  of\  the \ line}=-\frac 13\)
2x-2=-\(\frac 13\)
2x=-\(\frac 13\)+2
2x=\(\frac 53\)
x=\(\frac 56\)
Now, x=\(\frac 56\)
y=\(\frac {25}{36}-\frac {10}{6}+7\)= \(\frac {25-60+252}{36}\) = \(\frac {217}{36}\)
Thus, the equation of the tangent passing through(\(\frac 56\),\(\frac {217}{36}\)) is given by
y - \(\frac {217}{36}\) =-\(\frac 13(x-\frac 56)\)
\(\frac {36y-217}{36 }\)= -\(\frac {1}{18}\)(6x-5)
36y-217 = -2(6x-5)
36y-217 = -12x+10
36y+12x-227 = 0

Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y−15x =13) is 36y+12x-227=0.