Can someone help me to solve this : Find the equation of the tangent to the curve x^{2}-2xy + 2y^{2}-7x + 6y + 6 = 0 which is perpendicular to 6x + 5y -4 = 0

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- Jun 17th 2012, 12:56 PMmgmanojEquation of the tangent to a perpendicular lin
Can someone help me to solve this : Find the equation of the tangent to the curve x

^{2}-2xy + 2y^{2}-7x + 6y + 6 = 0 which is perpendicular to 6x + 5y -4 = 0 - Jun 17th 2012, 01:12 PMPlatoRe: Equation of the tangent to a perpendicular lin
- Jun 17th 2012, 01:30 PMmgmanojRe: Equation of the tangent to a perpendicular lin
After differentiating the equation I got

__dy__=__2x-2y-7__

dx 2x-4y-6

the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here. - Jun 17th 2012, 01:53 PMPlatoRe: Equation of the tangent to a perpendicular lin
- Jun 17th 2012, 05:18 PMHallsofIvyRe: Equation of the tangent to a perpendicular lin
The equation 6x+ 5y- 4= 0 is the same as y= -(6/5)x+ 4/6 and has slope -6/5. The "negative reciprocal" of that is 6/5 and is the slope of the perpendicular line. I don't know where you got "45/6"! Did you mean "4/6"? Remember that the

**slope**of the line, the number multiplying x, determines the direction, not the "constant" term.