The focal point of the ellipse X & # 178 / 25 + Y & # 178 / 16 = 1 is F & # 8321;, F & # 8322;, P is the point on the ellipse, if ∠ F & # 8321; PF & # 8322; = Π / 2, then the area of △ F & # 8321; PF & # 8322; is equal to

The focal point of the ellipse X & # 178 / 25 + Y & # 178 / 16 = 1 is F & # 8321;, F & # 8322;, P is the point on the ellipse, if ∠ F & # 8321; PF & # 8322; = Π / 2, then the area of △ F & # 8321; PF & # 8322; is equal to

When a = 5, B = 4, C = 3 ∠ f1pf2 = π / 2, the trajectory of point P is a circle with origin o (0,0) as the center and radius 3, and there is no intersection point between the circle and the original ellipse. If P is on the ellipse, ∠ f1pf2 < π / 2 is certain. It can't be equal to 90 degree

The focus of the ellipse X & # 178; / 9 + Y & # 178; / 2 = 1 is F & # 8321; F & # 8322;, and the point P is on the ellipse if pf & # 8321; = 4, PF & # 8322; = 2
The size of ∠ F &; PF &; is?

Using cosine theorem, | F1F2 | = 2 √ 7, cos ∠ F & # 8321; PF & # 8322; = (16 + 4 - 28) / (2 × 4 × 2) = - 1 / 2,
∴ ∠F₁PF₂= 120º.

If point O and point F are the center and left focus of the ellipse X & # 178 / 4 + Y & # 178 / 3 = 1 respectively, point P is any point on the ellipse
Want to get help: 1. The maximum value of the vector OP multiplied by the vector FP is?
How to deduce OP = (x, y) and FP = (x + 1, y) in the process of 2?

a=2,b=√3,c=1,O(0,0),F(-1,0)
1, let P (x, y), then vector OP = (x, y), vector FP = (x + 1, y)
So the vector OP * vector FP = x (x + 1) + Y & # 178; = x & # 178; + X + Y & # 178;
If P (x, y) is on the ellipse X & # 178 / 4 + Y & # 178 / 3 = 1, then y & # 178; = 3-3x & # 178 / 4
So the vector OP * vector FP = x & # 178; + X + 3-3x & # 178; / 4
=1/4*x²+x+3
=1/4*(x+2)²+2
If - 2 ≤ x ≤ 2, then when x = 2, the maximum value of 1 / 4 * (x + 2) & # 178; + 2 is 1 + 2 + 3 = 6
That is, the maximum value of the vector OP * vector FP is 6
2. The coordinates of the vector are the coordinates of the point of the arrow minus the coordinates of the point of the tail. Isn't that what the book says

It is known that m and N are the two ends of the major axis of the ellipse C, and the product of the slopes of PM and PN is - 3 / 4, then the eccentricity of the ellipse is -

I'll take it. It's 0.5

It is known that the coordinates of M and N are (- √ 2,0) and (√ 2,0), respectively. The lines PM and PN intersect at point P, and the product of their slopes is - 1 / 2
Finding the trajectory equation of point P

P(x,y)
(y-0)/(x+√2)/(y-0)/(x-√2)=-1/2
y²/(x²-2)=-1/2
x²-2=-2y²
X & # 178 / / 2-y & # 178; = 1 and Y ≠ 0

In RT △ ABC, ∠ C = 90 °, ab = 13, BC = 12, find Sina, cosa, Tana

Well, isn't that simple? It's a Pythagorean theorem
∵∠C=90° AB=13.BC=12
∴AC^2=AB^2-CB^2
So AC = 5
∴sinA=12/13 cosA=5/13 tanA=12/5
If you don't know, ask again,

In △ ABC, a ∶ B ∶ C = 5 ∶ 12 ∶ 13, then Sina=___ ,cosA=___

Sina is 5 / 13
Cosa is 12 / 13

Given the function f (x) = sin (x + 3 / 2) sin (x2), find the maximum value and the minimum positive period of F (x) {/ 2

f(x) = sin(x+3π/2)sin(x-2π)
= -cosxsinx
= -1/2sin2x
1 / 2 max
Minimum - 1 / 2
Minimum positive period 2 π / 2 = π
f(π/6)=-1/2sinπ/3=-√3/4
f(π/12)=-1/2sinπ/6=-1/4
f(π/6)+f(π/6)=-(√3+1)/4

The minimum positive period of the function f (x) = sin (x + π 3) sin (x + π 2) is t=______

Y = sin (x + π 3) sin (x + π 2) = (sinxcos π 3 + cosxsin π 3) cosx = 12sinxcosx + 32cos2x = 14sin2x + 32 · 1 + cos2x2 = 34 + 12sin (2x + π 3) ‖ t = π

Given that the minimum positive period of the function f (x) = sin (2 ω X - π / 3) (ω > 0) is π, find ω

The minimum positive period T = 2 π / 2W = π;
∴w=1;
If you don't understand this question, you can ask,