Answer:

The equation of the given curve is y=x

2

−2x+7.

On differentiating with respect to x, we get:

dx

dy

=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

.

⇒

dx

dy

=2x−2=

3

−1

⇒2x=

3

−1

+2

⇒2x=

3

5

⇒x=

6

5

Now, at x=

6

5

⇒y=

36

25

−

6

10

+7=

36

25−60+252

=

36

217

Thus, the equation of the tangent passing through (

6

5

,

36

217

) is given by,

(y−

36

217

)=−

3

1

(x−

6

5

)

⇒=

36

36y−217

−

18

1

(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

Answer:

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