Find the equation of equiaxed hyperbola which passes through point a (3, - 1) and whose symmetry axis is on the coordinate axis

Find the equation of equiaxed hyperbola which passes through point a (3, - 1) and whose symmetry axis is on the coordinate axis

When the focus is on the x-axis, let the standard equation of hyperbola be x2a2 − y2a2 = 1, substituting a (3, - 1) into the equation to get 9A2 − 1A2 = 1, A2 = 8, and the standard equation of hyperbola be x28 − Y28 = 1
It is proved that (sin α + sin β) / (COS α - cos β) = cot [(β - α) / 2]
I only prove Tan [(α + β) / 2]
Using the sum difference product formula,
=[2sin((α+β)/2)·cos((α-β)/2)]/[2sin((α+β)/2)·sin((α-β)/2)]
=cot [(β-α)/2]
Find the equation of equiaxed hyperbola which passes through point a (3, - 1) and whose symmetry axis is on the coordinate axis
When the focus is on the x-axis, let the standard equation of hyperbola be x2a2 − y2a2 = 1, substituting a (3, - 1) into the equation to get 9A2 − 1A2 = 1, A2 = 8, and the standard equation of hyperbola be x28 − Y28 = 1
Given Tan α = 2, find the value of (1 + 2Sin α cos α) / (Sin & sup2; α - cos & sup2; α)
I'm not good at math
(1+2sinαcosα）/（sin²α-cos²α）
=(sina+cosa)^2/(sina+cosa)(sina-cosa)
=(sina+cosa)/(sina-cosa)
The numerator and denominator are divided by cosa
=(tana+1)/(tana-1)
=3
16/3
sin²α+cos²α=1
The original formula = (Sin & sup2; α + cos & sup2; α + 2Sin α cos α) / (Sin & sup2; α - cos & sup2; α)
=（sinα+cosα)²/[（sinα+cosα）（sinα-cosα）]
=（sinα+cosα）/（sinα-cosα）
Division of cos α
=（tanα+1）/（tanα-1）
=3
Find through the point a (3,1), and the axis of symmetry are on the axis of the equiaxed hyperbola equation!
An equiaxed hyperbola whose axis of symmetry is on the coordinate axis
x^2/a^2 -y^2/a^2 =1
Point (a, - 1)
9/a^2 - 1/a^2 =1
a^2=8
x^2-y^2=8
So the equation is: x ^ 2 / 8-y ^ 2 / 8 = 1
3^2/a^2-1^2/a^2=1
3^2/a^2-1^2/a^2=1
9/a^2-1/a^2=1
8/a^2=1
a^2=8
x^2/8-y^2/8=1
Let x ^ 2 / A ^ 2 - y ^ 2 / A ^ 2 = 1 be the equiaxed hyperbola with all symmetry axes on the coordinate axis
Because it passes through point a (3, - 1), substituting
9/a^2 - 1/a^2 =1
We get a ^ 2 = 8
x^2-y^2=8
So ^ 2 / y = 1 / 8
Given that Tan (π / 4 + θ) + Tan (π / 4 - θ) = 4, and - π < θ < - π / 2, find the value of Sin & sup2; θ - 2Sin θ cos θ - cos & sup2; θ
tan( π/4 + θ ) + tan( π/4 - θ ) = 4
[ sin( π/4 + θ )cos( π/4 - θ ) + cos( π/4 + θ )sin( π/4 - θ )] / [cos( π/4 + θ )cos( π/4 - θ )] = 4
sin( π/2 ) / [cos( π/4 + θ )cos( π/4 - θ )] = 4
cos( π/4 + θ )cos( π/4 - θ ) = 1 / 4
( cos π/2 + cos 2θ ) / 2= 1 / 4
cos 2θ = 1 / 2
sin 2θ = √( 1 - cos² 2θ ) = √3 / 2
sin² θ - 2sin θ cos θ - cos² θ = - ( sin 2θ + cos 2θ ) = - (√3 + 1) / 2
The equation of an equiaxed hyperbola passing through the point P (3, - 1) with its axis of symmetry on the coordinate axis is ()
A. x2-y2=10B. y2-x2=10C. x2-y2=8D. y2-x2=8
Let the hyperbolic equation be x2a2 − y2a2 = 1 (a > 0), then through a (3, - 1), A2 = 8 to get x28 − Y28 = 1, that is, x2-y2 = 8, so choose C
It is known that 2sin2 α + 2Sin α cos α 1 + Tan α = K (0 ＜ α＜ π 2)
2sin2 α + 2Sin α cos α 1 + Tan α = 2Sin α (sin α + cos α) 1 + sin α cos α = 2Sin α cos α (sin α + cos α) sin α + cos α = 2Sin α cos α = K
Find the equation of equiaxed hyperbola which passes through point a (3, - 1) and whose symmetry axis is on the coordinate axis
When the focus is on the x-axis, let the standard equation of hyperbola be x2a2 − y2a2 = 1, substitute a (3, - 1) into the equation to get 9A2 − 1A2 = 1, A2 = 8, and the standard equation of hyperbola be x28 − Y28 = 1; (4 points) when the focus is on the y-axis, let the standard equation of hyperbola be y2a2 − x2a2 = 1, and substitute a (3, - 1) into the equation to get 1A2 − 9A2 = 1, A2 = - 8, which does not exist
Given (1 + Tan θ) / (1-tan θ) = 3 + 2 √ 2, find 1 / 2 Sin & sup2; θ - sin θ cos θ
(1 + Tan θ) / (1-tan θ) = 3 + 2 √ 2tan θ = √ 2 / 2, so sin θ / cos θ = √ 2 / 2cos θ = √ 2Sin θ cos & sup2; θ = 2Sin & sup2; θ = 1, so Sin & sup2; θ = 1 / 5sin θ cos θ = sin θ * √ 2Sin θ = √ 2Sin & sup2; θ = √ 2 / 5, so the original formula = (...)