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What is the equation of the tangent at the vertex of this parabola? $$4y^2+6x=8y+7$$

I simplified the equation and got $$4(y-1)^2 =-(6x-11)$$ What do I do further?

Blue
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Bismeet singh
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  • The parabola $y^2=x$ has its vertex at $(0,0)$. You could see your parabola as a translated standard parabola, what has the vertex translated to? Next, you should use explicit differentiation to find the slope at the vertex and set up the equation for the tangent line. – rae306 Apr 30 '16 at 08:54
  • So I have to differentite the equation found after simplification? – Bismeet singh Apr 30 '16 at 09:00
  • Yes, you can isolate $y$ and then differentiate OR use implicit differentiation (don't know if you have learned that yet) and differentiate the original equation. – rae306 Apr 30 '16 at 09:01
  • Differentiate the original equation or the one I got after simplification? – Bismeet singh Apr 30 '16 at 09:02
  • It is $y-1=0$.. – Archis Welankar Apr 30 '16 at 09:09

2 Answers2

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The equation of your parabola is $-4(y-1)^2=6(x-\tfrac{11}{6})$, therefore the vertex is $V(\tfrac{11}{6},1)$.

Since the parabola is upside down (opening to the left), the equation of the tangent line at the vertex is $x=\tfrac{11}{6}$.

rae306
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Hint. Consider $x$ to be a function of $y.$ Then we must have that $$11-6x=4(y-1)^2$$ increases indefinitely as $y$ increases. Thus there is a minimum value where the nonnegative quantity $4(y-1)^2$ vanishes. This is the vertex of your parabola, so that we must have that $11-6x=0$ at the vertex. Can you now find the coordinates of this turning point?

Allawonder
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