Question

Find the equation of the tangent line to the curve y = x² − 2x + 7 which perpendicular to the line 5y − 15x = 13. ?

Answer 1

The equation of the given curve is `y=x^2-2x+7`

On differentiating with respect `x,` we get:

`(dy)/(dx)=2x-2`

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

`5y-15x=13`

`rArr5y=15x+13`

`rArr(5y)/5=(15x)/5+13/5` .....................[dividing both the sides by 5]

`rArry=3x+13/5`

This is of the form y = mx + c

`therefore" slope of the line = 3"`

If a tangent is perpendicular to the line 5y - 15x = 13, then the slope of the tangent is `(-1)/("slope of the line")=(-1)/3`

`rArr2x-2=(-1)/3`

`rArr2x=(-1)/3+2`

`rArr2x=(-1+6)/3`

`rArr2x=5/3`

`rArrx=5/(3xx2)`

`rArrx=5/6`

Now, `x = 5/6`

`rArry=x^2-2x+7`

`rArry=(5/6)^2-2(5/6)+7`

`rArry=25/36-10/6+7`

`rArry=25/36-60/36+252/36`

`rArry=(25-60+252)/36=217/36`

Thus, the equation of the tangent passing through `(5/6, 217/36)`

`y-y_1=m(x-x_1)`

`rArry-217/36=-1/3(x-5/6)`

`rArr(36y-217)/36=-1/18(6x-5)`

`rArr36y-217=-2(6x-5)`

`rArr36y-217=-12x+10`

`rArr36y+12x-217-10=0`

`rArr36y+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