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(a) parallel to the line 2x−y+9=0.

(b) perpendicular to the line 5y−15x=13.

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Solution

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(a)

The equation of the given curve is y=x2−2x+7.On differentiating with respect to x, we get:

dydx=2x−2

The equation of the line is 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, at x=2

⇒y=22−2×2+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.

(b)

The equation of the line is 5y−15x=13.

Slope of the line =3

If a tangent is perpendicular to the line 5y−15x=13,

then the slope of the tangent is −1Slope of the line=−13.

⇒dydx=2x−2=−13

⇒2x=−13+2

⇒2x=53

⇒x=56

Now, at x=56

⇒y=2536−106+7=25−60+25236=21736

Thus, the equation of the tangent passing through (56,21736) is given by,

(y−21736)=−13(x−56)

⇒=36y−21736−118(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

Slope of the line =3

If a tangent is perpendicular to the line 5y−15x=13,

then the slope of the tangent is −1Slope of the line=−13.

⇒dydx=2x−2=−13

⇒2x=−13+2

⇒2x=53

⇒x=56

Now, at x=56

⇒y=2536−106+7=25−60+25236=21736

Thus, the equation of the tangent passing through (56,21736) is given by,

(y−21736)=−13(x−56)

⇒=36y−21736−118(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

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