Find a point in the ellipse C: x ^ 2 + 4Y ^ 2 = 16. Find the equation of the line where the chord is located with a point a (1, - 1) in the ellipse x * 2 + 4Y * 2 = 16 as the midpoint

Find a point in the ellipse C: x ^ 2 + 4Y ^ 2 = 16. Find the equation of the line where the chord is located with a point a (1, - 1) in the ellipse x * 2 + 4Y * 2 = 16 as the midpoint

Let slope + Veda theorem
Through a: y = K (x-1) - 1
(4K * 2 + 1) x * 2-8k (K + 1) x + 4 (k * 2 + 1) * 2-16 = 0
X1+x2=8k(k+1)/(4k*2+1)=2
K=1/4

In the ellipse x + 4Y = 16, find the line equation of the chord passing through the point m (2,1) and bisected by this point Chord length

Let the linear equation be y = K (X-2) + 1 simultaneous x ^ 2 + 4Y ^ 2 = 16, then (1 + 4K ^ 2) x ^ 2 - (16K ^ 2-8k) x + (16K ^ 2-16k-12) = 0
Let the intersection point of line and ellipse be a (x1, Y1); B (X2, Y2) X1 + x2 = (16K ^ 2-8k) / (1 + 4K ^ 2)
(x1 + x2) / 2 = (16K ^ 2-8k) / 2 (1 + 4K ^ 2) = 2, k = - 1 / 2, so the linear equation is y = - 1 / 2 (X-2) + 1, that is, x + 2y-4 = 0

Find the equation of the line where the chord is located with a point a (1, - 1) in the ellipse x2 + 4y2 = 16 as the midpoint

Let the chord of an ellipse intersect with the ellipse at the midpoint of a (1, - 1) at e (x1, Y1), f (X2, Y2),
∵ a (1, - 1) is the midpoint of EF,
∴x1+x2=2，y1+y2=-2，
Replace e (x1, Y1), f (X2, Y2) into the ellipse x2 + 4y2 = 16, respectively,
have to
x12+4y12＝16
x22+4y22＝16 ，
∴（x1+x2）（x1-x2）+4（y1+y2）（y1-y2）=0，
∴2（x1-x2）-8（y1-y2）=0，
∴k=y1−y2
x1−x2=1
4，
The linear equation of the chord of an ellipse with a (1, - 1) as the midpoint is Y - (- 1) = 1
4（x-1），
After finishing, x-4y-5 = 0

If P (2, - 1) is the midpoint of chord ab of circle (x-1) ^ 2 + y ^ 2 = 25, what is the equation of line AB?

Point O (1,0) at the center of the circle connects op. the line OP is perpendicular to ab. the equation of OP is y = - x + 1;
So AB's equation is y = x-3

If P (2, - 1) is the midpoint of chord ab of circle (x-1) 2 + y2 = 25, then the equation of line AB is () A. x-y-3=0 B. 2x+y-3=0 C. x+y-1=0 D. 2x-y-5=0

The center of the circle is known to be o (1, 0)
Kop = 0 + 1
1−2＝−1
kABkOP=-1
K AB = 1, and the straight line AB crosses the point P (2, - 1),
The equation of line AB is x-y-3 = 0
So choose a

I'm sorry to trouble you if P (2, - 1) is the square of circle (X -- 1) and the square of Y is equal to the midpoint of chord ab of 25 to find the equation sum of line ab Find the equation and chord length ab of line ab

Let the slope of a (x1, Y1) B (X2, Y2) AB be K ∵ a, and B are points on the circle

There is a point P (- 1,2) in the circle x square + y square = 8. AB is the chord passing through the point P. when AB is the shortest time, find the equation of the straight line ab Come on, thank you

Oh, the shortest time when AB is perpendicular to radius
Give me points, I want to score!

If point P (3, - 1) is the midpoint of chord ab of circle (X-2) 2 + y2 = 25, then the equation of line AB is______ .

From (X-2) 2 + y2 = 25,
C (2, 0)
∴kPC＝0+1
2−3＝−1．
∵PC⊥AB，
∴kAB=1．
The equation of line AB is
y+1=x-3．
That is, X-Y-4 = 0
So the answer is: X-Y-4 = 0

If P (2,1) is the midpoint of chord ab of circle (x-1) + y square = 25, then the equation of line AB is given The slope of PC is (1-0) / (2-1) = 1 Then the slope is - 1 Why is the slope of ab - 1

According to the vertical longitude theorem, AB must be perpendicular to CP to find out CP: y = x-1kcp = 1. Because it is vertical, KCP * KAB = - 1 (this is the theorem. After you learn vector, you can use vector point multiplication formula to prove it, but in the end it is all formulas, so you may as well

If the chord ab of circle x * 2 + y * 2 = 20 is made through point P (2, - 3), and P is bisected into AB, then the equation of the line where chord AB is located is

The center of the circle is the origin o
P bisection ab
So AB vertical Po
The Po slope is - 3 / 2
So the slope of AB is 2 / 3
Over P
So 2x-3y-13 = 0