Mathematics, 27.07.2019 05:00 Goldenstate32

The equation of a hyperbola is x2 − 4y2 − 2x − 15 = 0. the width the asymptote rectangle is units, and its height is units.

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The equation of the hyperbola given is not in standard form. Before doing this problem we need to write the equation in standard form. The standard form for a hyperbola is: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}$

Before we get into what the h, k, a and b represent, let’s get the equation we are given into standard form.

First we pull aside the terms that have x in them. These are $x^{2} -2x$ . We wish to write these as a perfect square but as they are, they are not a perfect square. Notice that if we add 1 we would get a perfect square. That is, $x^{2} -2x+1=(x-1)^2$ . While this is a perfect square we have added 1 to the terms originally given. This changes the equation and so to keep it balanced we add 1 to the other side as well. We get this:

$( x-1)^{2} - 4y^2-15=0+1$

This is still not in standard form. Let’s move the constant term (the number -15) to the right hand side. We get: (x-1)^2-4y^2=15+1

Notice that the equation in standard form is equal to 1 but ours is equal to 16. So we divide every term by 16 and this yields:

$\frac{(x-1)^2}{16}- \frac{y^2}{4}=1$

It is almost in standard form. In standard form the denominators are written $a^{2}$ and $b^{2}$ so we take the numbers in the denominators and write 4 squared instead of 16 and 2 squared instead of 4.

This finally gives the equation in standard form. It is:

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

Comparing this to the standard form we see h=1 and k=0. This is the center of the hyperbola (h,k) which is (1,0).

The denominator of the first fraction gives the value a which here is 4. Twice this value is the length of the rectangle used to find the asymptotes. That is, the length of the asymptotic rectangle is 8.

The denominator of the second fractions gives the value b which here is 2. Twice this value is the width of the rectangle used to find the asymptotes. that is, the width of the asymptotic rectangle is 4.

Though you are not asked to graph this hyperbola. The diagonals of the rectangle would help us draw it as they would give the asymptotes.

Answer from: Quest

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

In the diagram of circle o, what is the measure of abc? 210 150

Answer from: Quest

The answer is: y would equal 3 because if you simplify the equation then it would be y= 3x+9 and if you divide 3x and 9 by 3 it would equal 3. y= 3

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The equation of a hyperbola is x2 − 4y2 − 2x − 15 = 0. the width the asymptote rectangle is units,...