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Question
Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is perpendicular to the line 5y − 15x = 13.
Solution
`The equation of the given curve is y = x2 − 2x + 7
On differentiating with respect to x, we get:
`"dy"/"dx" = 2x - 2`
The equation of the line is 5y − 15x = 13.
5y − 15x = 13
⇒ `"y" = 3x + 13/5`
This is of the form y = mx + c.
∴ 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`
⇒ 2x - 2 = `(-1)/3`
⇒ 2x = `(-1)/3 + 2`
⇒ 2x = `5/3`
⇒ x = `5/6`
Now, x = `5/6`
`=> "y" = 25/36 - 10/6 + 7 = (25 - 60 + 252)/36 = 217/36`
Thus, the equation of the tangent passing through `(5/6, 217/36)` is given by,
`"y" - 217/36 = -1/3 (x - 5/6)`
`=> (36"y" - 217)/36 = (- 1)/18 (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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