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# Find the equation of the tangent line to the curve y=x2−2x+7 which is.(a) parallel to the line 2x−y+9=0.(b) perpendicular to the line 5y−15x=13.

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−2The equation of the line is 2x−y+9=0.⇒y=2x+9This is of the form y=mx+c.Slope of the line =2If 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=2Now, at x=2⇒y=22−2×2+7=7Thus, the equation of the tangent passing through (2,7) is given by, y−7=2(x−2)⇒y−2x−3=0Hence, 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 =3If 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=56Now, at x=56⇒y=2536−106+7=25−60+25236=21736Thus, 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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Similar Questions
Q1
Find the equation of the tangent line to the curve y=x22x+7 which is.
(a) parallel to the line 2xy+9=0.
(b) perpendicular to the line 5y15x=13.
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Q2
Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is (i) parallel to the line 2x − y + 9 = 0 (ii) perpendicular to the line 5y − 15x = 13.
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Q3
Find the equation of the tangent line to the curve y=x22x+7 which is perpendicular to the line 5y15x=13.
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Q4

Find the equation of the tangent line to the curve y=x22x+7 which is

(a) parallel to the line 2x-y+9=0.

(a) parallel to the line 5y-15x=13.

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Q5
The equation of the tangent line to the curve y=x22x+7 which is perpendicular to the line 5y15x=13 is 12x+36y227=0.If true enter 1 else 0.
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