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Lesechka [4]
3 years ago
13

Find the two horizontal and two vertical tangent lines of the equation x^2+xy+y^2=9.

Mathematics
1 answer:
valina [46]3 years ago
3 0
Differentiating w.r.t y,
We get, 

<span>2 x (d x/d y) + y (d x/d y) + x + 2 y = 0 </span>

<span>Put 
d x/d y = 0 </span>

The equation becomes:
<span>x=2 y </span>

<span>substituting into main equation

We, get

3 y^2 =9 </span>

y= √3

x = <span>± 2√3</span>
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Which is the equation of a hyperbola with directrices at y = ±2 and foci at (0, 4) and (0, −4)?
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Answer:

Step-by-step explanation:

First off, I'm assuming that when you said "directrices" you mean the oblique asymptotes, since hyperbolas do not have directrices they have oblique asymptotes.

If we plot the asymptotes and the foci, we see that where the asymptotes cross is at the origin. This means that the center of the hyperbola is (0, 0), which is important to know.

After we plot the foci, we see that they are one the y-axis, which is a vertical axis, which means that the hyperbola opens up and down instead of sideways. Knowing those 2 characteristics, we can determine that the equation we are trying to fill in has the standard form

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1

We know h and k from the center, now we need to find a and b. Those values can be found from the asymptotes. The asymptotes have the standard form

y = ±\frac{a}{b}(x-h)+k

Filling in our asymptotes as they were given to us:

y = ±\frac{2}{1}(x-0)+0 where a is 2 and b is 1. Now we can write the formula for the hyperbola!:

\frac{(y-0)^2}{4}-\frac{(x-0)^2}{1}=1 which of course simplifies to

\frac{y^2}{4}-\frac{x^2}{1}=1

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Thank you for posting your question here at brainly. Feel free to ask more questions.  

<span>The best and most correct answer among the choices provided by the question is 86.</span><span>       
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<span>Hope my answer would be a great help for you.</span></span>
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