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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Find vertical tangent

• Thread starter Stuckonproblems
• Start date Apr 12, 2012

S

Stuckonproblems

New member

Joined Apr 10, 2012 Messages 7 Find the point(s) on the curve at which the tangent line is(are) vertical given the curve x2+xy+y2​=3 G

galactus

Super Moderator

Staff member Joined Sep 28, 2005 Messages 7,203 Differentiate implicitly: \(\displaystyle y'=\frac{-(2x+y)}{x+2y}\) Now, vertical tangents occur where the slope is undefined. When the denominator is 0. Set the denominator equal to 0 and solve for y. Sub this back into the original and solve for the x. y coordinates follow. S

Stuckonproblems

New member

Joined Apr 10, 2012 Messages 7 So I set the denominator equal to 0. x+2y= 0 x=-2y y=-x/2 so I plug these two back into the original equation x2+xy+y2=3 to find the x and y points for the tangent points right? [FONT=MathJax_Math]y[/FONT]​ H

HallsofIvy

Elite Member

Joined Jan 27, 2012 Messages 7,760 Those are not two separate equations. Putting y= -x/2 into \(\displaystyle x^2+ xy+ y^2= 3\) gives \(\displaystyle x^2- x^2/2+ x^2/4=3x^2/4= 3\). Solve that for x and then use y= -x/2 to find the corresponding values for y. OR put x= -2y into the equation: \(\displaystyle 4y^2- 2y^2+ y^2= 3y^2= 3\). Solve that for y and then use x= -2y to find the correspondinmg values for x. You must log in or register to reply here. Share: Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link

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