Question

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

Answer 1

The equation of the given curve is,

y= x 2 −2x+7

The slope of the tangent to the curve is given by,

slope= dy dx

Hence, the slope of the tangent to the given curve is,

dy dx = d( x 2 −2x+7 ) dx =2x−2 (1)

The equation of the line is given by,

2x−y+9=0 y=2x+9

The above equation is written in the form of y=mx+c , where m is the slope of the line. Hence, slope of the given line is 2.

For a tangent to be parallel to the line, their slopes must be equal. Therefore,

2x−2=2 x=2

Coordinate of y when x=2 is,

y=4−4+7 =7

Thus, the tangent passes through the point ( 2,7 ).

Equation of the line passing through a point ( x 1 , y 1 ) is given by,

y− y 1 =m( x− x 1 )(2)

Hence, equation of the tangent through point ( 2,7 )and slope m=2 is,

y−7=2( x−2 ) y−2x−3=0

Hence, the equation of the curve parallel to 2x−y+9=0 is y−2x−3=0.

The equation of the given line is,

5y−15x=13 y=3x+ 13 5

The above equation is written in the form of y=mx+c, where, m is the slope of the line. Hence, slope of the given line is 3.

When the tangent is perpendicular to the line 5y−15x=13, the slope of the tangent will be,

−1 slope of the line = −1 3

Hence, the slope of the tangent from equation (1) is,

2x−2= −1 3 x= 5 6

Coordinate of y when x= 5 6 is,

y=3( 5 6 )+ 13 5 = 217 36

Thus, the tangent passes through the point ( 5 6 , 217 36 ).

Hence, equation of the tangent through point ( 5 6 , 217 36 )and slope m= −1 3 from equation (2) is,

y− 217 36 = −1 3 ( x− 5 6 ) 36y+12x−227=0

Thus, the equation of the tangent to the curve perpendicular to the line 5y−15x=13 is 36y+12x−227=0.