Topic source: homework help

Known ellipse C: X2A2+Y2B2 =1(a >; B>0) has an eccentricity of 2√2, a circle with the origin as the center and the short semi-axis length of ellipse C as the radius, and a straight line X? Y+2√=0 tangent.

(1) Find the equation of ellipse c;

(2) If the straight line passing through point M (2,0) intersects ellipse C and points A and B, and O is the coordinate origin, is there a point P on ellipse C, which makes quadrilateral OAPB a parallelogram? Please provide a justification for the answer.

Test center/site

Simple properties of ellipse

analyse

(i) Using the distance formula from point to straight line and eccentricity formula, the values of a and b can be obtained, and the equation of ellipse c can be obtained;

(Ⅱ) Set the equation of straight line L, substitute it into elliptic equation, and use Vieta theorem and

OA+

OB=

OP, you can get the value of k immediately, K2 = 1 14.

explain

(1) b = 2 √1+1√ =1,E2 = C2A2 = A2? b2a2= 12，

∴a2=2,b2= 1

The equation of ellipse c is X22+Y2 = 1...(5 points)

(2) From the meaning of the question, we know that the slope of a straight line exists. Let the slope of a straight line be k, and its equation is y=k(x? 2),

By who? y=k(x？ 2)x22+y2= 1, and get (1+2k2)x2? 8k2x+8k2？ 2 = 0 ...(6 points)

∫△= 64k 4？ 4(8k2？ 2)( 1+2k2)>0,

∴k2<； 12 ...(7 points)

Let A(x 1, y 1), B(x2, y2), and suppose there is a point P(x, y) on ellipse C, so that the quadrilateral OAPB is a parallelogram.

Then there are x 1+x2 = 8k2 1+2k2, x 1? x2=8k2？ 2 1+2k2，OA？ →? +OB？ →? =OP？ →? ,

∴(x 1,y 1)+(x2,y2)=(x,y)

∴x=x 1+x2=8k2 1+2k2,y=y 1+y2=k(x 1？ 2)+k(x2？ 2)=? 4k 1+2k2

Point P(x, y) is on ellipse c,

∴(8k2 1+2k2)2+2(？ 4k 1+2k2)2=2, which means 28k4+ 12k2? 1=0

Solution: K2 = 1 14

So there is a point p on ellipse C, which makes quadrilateral OAPB a parallelogram ... (12 minutes)