Question

(a) parallel to the line $2x−y+9=0$.

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

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Updated on : 2022-09-05

Solution

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(a)

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

$dxdy =2x−2$

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 $x=2$

$⇒y=2_{2}−2×2+7=7$

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 $Slope of the line−1 =3−1 $.

$⇒dxdy =2x−2=3−1 $

$⇒2x=3−1 +2$

$⇒2x=35 $

$⇒x=65 $

Now, at $x=65 $

$⇒y=3625 −610 +7=3625−60+252 =36217 $

Thus, the equation of the tangent passing through $(65 ,36217 )$ is given by,

$(y−36217 )=−31 (x−65 )$

$⇒=3636y−217 −181 (6x−5)$

$⇒36y−217=−2(6x−5)$

$⇒36y−217=−12x+10$

$⇒36y+12x−227=0$

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 $Slope of the line−1 =3−1 $.

$⇒dxdy =2x−2=3−1 $

$⇒2x=3−1 +2$

$⇒2x=35 $

$⇒x=65 $

Now, at $x=65 $

$⇒y=3625 −610 +7=3625−60+252 =36217 $

Thus, the equation of the tangent passing through $(65 ,36217 )$ is given by,

$(y−36217 )=−31 (x−65 )$

$⇒=3636y−217 −181 (6x−5)$

$⇒36y−217=−2(6x−5)$

$⇒36y−217=−12x+10$

$⇒36y+12x−227=0$

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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