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I need to find the equation of the tangent to the hyperbola $$\frac{x^2}{6}-\frac{y^2}{8}=1$$ at the point $(3,2)$.

I tried doing it by substituting for $y$ but the algebra is not nice at all and I wanted to see how people here would do it.

Sean B
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  • you're missing the minus sign. – hjhjhj57 Jan 26 '15 at 22:53
  • Have you done any implicit differentiation? – turkeyhundt Jan 26 '15 at 22:54
  • Yeah sorry that has been corrected. And I have done implicit differentiation but I'd rather solve this without calculus if possible. – Sean B Jan 26 '15 at 22:55
  • @SeanB: Without calculus at all? That won't be easy. – hmakholm left over Monica Jan 26 '15 at 22:56
  • I know how to calculate it with calculus but I would like to do it elsewhere. Bonus question: when you calculate dy/dx it is in terms of both x and y so you need both values of x and y to calculate the gradient at that point correct? – Sean B Jan 26 '15 at 23:01
  • Yes, in this case you need both the $x$ and $y$ to calculate dy/dx – turkeyhundt Jan 26 '15 at 23:02
  • You could set up a system using a line going through the point (3,2) with an undetermined slope m and the equation for the hyperbola. Solve the system, which will leave you with a quadratic. The discriminant for this quadratic must be 0, why? From here, you should be able to obtain a value for m. – Winston Jan 26 '15 at 23:05
  • @Winston I see that D must be 0 as this otherwise there will be two roots and then it would not be a tangent but the algebra is absolutely horrible and I always make mistakes. Any advice? – Sean B Jan 26 '15 at 23:07
  • Sean, I agree the algebra is messy. Wolfram alpha may help! – Winston Jan 26 '15 at 23:10
  • That would be a good start but I was thinking about an exam type situation also. Guess I will try and stick to calculus techniques. Thanks. – Sean B Jan 26 '15 at 23:11
  • ugh, I hope a teacher isn't so cruel that he or she would make you do all that algebra. – Winston Jan 26 '15 at 23:14

2 Answers2

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If would compute the gradient of $x^2/6-y^2/8$ at the given point. If nonzero (which it is), it is perpendicular to the level set.

hmakholm left over Monica
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you don't need to solve for $y$. you can implicitly differentiate $x^2/6 - y^2/8 = 1$ to get $$\dfrac{xdx}{3} - \dfrac{ydy}{4} = 0$$ substitute $x = 3, y = 2$ to find the tangent has slope $$dx - \frac{dy}{2} = 0$$ turn this into the tangent by replacing $dx, dy$ by $(x-3), (y-2)$ so that the equation of the tangent line at (3,2) is $$(x-3) - \frac{(y-2)}{2}= 0.$$

abel
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