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The equation of a curve is $$3x^2 + 8xy + y^2 = -13.$$ Find the equations of the two tangents which are parallel to the $y$-axis.

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What have you tried? Are you aware of how to find a tangent at some point? –  Karolis Juodelė Aug 9 '12 at 13:15
    
You've tagged the problem "implicit differentiation." Do you know how to do implicit differentiation? Or is that the bit you can't do? –  Gerry Myerson Aug 9 '12 at 13:18
    
Well I managed to find dy/dx in terms of x and y, but what do i do from there? dy/dx is a tangent, isn't it? –  Tharun Aug 9 '12 at 13:22
    
$dy/dx$ is the slope of the tangent line. You're looking for places where that slope is infinite. This will probably correspond to points where your formula for $dy/dx$ results in dividing by zero. –  Gerry Myerson Aug 9 '12 at 23:59
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2 Answers 2

Hints :

  • differentiate (you should get a $f(x,y)dx+g(x,y)dy=0$ equation)
  • search the zeros of $\frac {dx}{dy}$ (you should get $y$ in function of $x$)
  • replace $y$ (for example) in the initial equation and solve
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The equation of any straight line parallel to the y-axis is x=c (where c is some constant).

If this line is tangent to the given curve, the point (c,y) must lie on both the line as well as on the curve, so, $3c^2+8cy+y^2=-13$

$=>y^2+y(8c)+(3c^2+13)=0$, but this is a quadratic equation in y.

As the line is tangent to the given curve, both the value of y should be same($\frac{-8c}{2}=-4c$) to make the two point of intersection coincide.

=>$(8c)^2=4(3c^2+13)$ as the discriminant needs to 0.

$=>c=±1$, so the equation of the required tangents are x=±1.

The point of contact being (c , -4c) ≡ (±1 , ∓4).

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