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.
Tangents to the curve and parallel to the $y$-axis?
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calculus
derivatives
implicit-differentiation
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0$dy/dx$ 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. – 2012-08-09
2 Answers
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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).