Question

Will give brainliest
Identify the type, center, and foci of the conic section. 11x2 – 2y2 + 66x – 16y + 45 = 0
a.ellipse with center (–3, 4), foci at (–3, 4 ± sqrt13)
b.hyperbola with center (3, 4), foci at (4 ±sqrt13, 3)
c. hyperbola with center (–3, –4), foci at (–3 ± sqrt13, –4)
ellipse with center (3, –4), foci at (3, –4 ±sqrt13 )

Answer 1

Answer:

Correct choice is C.

Step-by-step explanation:

The conic section is represented by equation

$11x^2-2y^2 + 66x-16y + 45 = 0.$

First, simplify this equation:

$(11x^2+66x)+(-2y^2-16y) + 45 = 0,\\ \\11(x^2+6x)-2(y^2+8y)+45=0,\\ \\11(x^2+6x+9-9)-2(y^2+8y+16-16)+45=0,\\ \\11((x+3)^2-9)-2((y+4)^2-16)+45=0,\\ \\11(x+3)^2-2(y+4)^2-99+32+45=0,\\ \\11(x+3)^2-2(y+4)^2=22,\\ \\\dfrac{(x+3)^2}{2}-\dfrac{(y+4)^2}{11}=1.$

This is the equation of hyperbola with center at point (-3,-4), real semiaxis $a=\sqrt{2}$ and imaginary semiaxis $b=\sqrt{11}.$ Then the distance from the center to the focus is

$c=\sqrt{a^2+b^2}=\sqrt{(\sqrt{2})^2+(\sqrt{11})^2}=\sqrt{2+11}=\sqrt{13}.$

Then foci are placed at points $(-3-\sqrt{13},-4),\ (-3+\sqrt{13},-4).$

Thus, correct choice is C.

Answer 2

Answer:

the answer is c

Step-by-step explanation:

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