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