Question

Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?
x squared over 16 minus y squared over 4 equals 1
x squared over 12 minus y squared over 4 equals 1
y squared over 16 minus x squared over 4 equals 1
y squared over 12 minus x squared over 4 equals 1

Answer 1

Answer:A. x squared over 16 minus y squared over 4 ...

Step-by-step explanation:

Answer 2

The correct option for the given directrix and foci of hyperbola is (B)

x squared over 12 minus y squared over 4 equals 1

Hyperbola:

The locus of a point in a plane which has two parts that are connected components are mirror image to one another with some conditions

How to find general equation of hyperbola?

Consider the general equation of hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Now we have to calculate the constants a and b by given directrix and foci

Directrix = x = ±3

$\frac{\pm a^2}{\sqrt{a^2+b^2}}=\pm 3\\\frac{ a^2}{\sqrt{a^2+b^2}}= 3\\\sqrt{a^2+b^2}=\frac{a^2}{3}$

Also we have foci = (4, 0)

$(\sqrt{a^2+b^2},0)=(4,0)\\\sqrt{a^2+b^2}=4$

On simplify these two equation we have

$\frac{a^2}{3}=4\\ \\\\a^2=12$

Substitute this value in above equation we have

$b^2=4$

Hence the required equation of the hyperbola is

$\frac{x^2}{12}-\frac{y^2}{4}=1$

The correct option is (B).

Learn more about Hyperbola here-

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