Question

Solution

Correct option is

2x2 + y2 = 1

Here Replacing (x, y) by  Then 3x2 + 3y2 + 2xy = 2 becomes ⇒                                       3(2x2 + 2y2) + 2(x2 – y2) = 4

⇒                                                               8x2 + 4y2 = 4

⇒                                                                 2x2 + y2 = 1

Which is free from the term containing xy.

SIMILAR QUESTIONS

Q1

O(0, 0), P(–2, –2) and Q(1, –2) are the vertices of a triangle, R is a point on PQ such that PR : RQ Q2

If three vertices of a rectangular are (0, 0), (a, 0) and (0, b), length of each diagonal is 5 and the perimeter 14, then the area of the rectangle is

Q3

If the line joining the points A(a2, 1) and B(b2, 1) is divides in the ratio b : a at the pint P whose x-coordinate is 7, their

Q4

If two vertices of a triangle are (3, –5) and (–7, 8) and centroid lies at the pint (–1, 1), third vertex of the triangle is at the point (a, b) then

Q5

α is root of the equation x2 – 5x + 6 = 0 and β is a root of the equation x2– x – 30 = 0, then coordinates of the point P farthest from the origin are

Q6 are two points whose mid-point is at the origin. is a point on the plane whose distance from the origin is

Q7

Locus of the point P(2t2 + 2, 4t + 3), where t is a variable is

Q8

If the coordinates of An are (n, n2) and the ordinate of the center of mean position of the points A1A2, … An is 46, then n is equal to

Q9

Area of the triangle with vertices A(3, 7), B(–5, 2) and C(2, 5) is denoted by Δ. If ΔA, ΔBΔC denote the areas of the triangle with vertices OBC, AOC and ABO respectively, O being the origin, then

Q10

A1A2A3, …. An are points in a plane whose coordinates are (x1y1), (x2y2), (x3y3) …, (xnyn) respectively

A1Ais dissected at the point G1GA3 is divided in the ratio  1 : 2 at G2GA4 is divided in the ratio 1 : 4 at G4, and so on until all n points are exhausted. The coordinates of the final point G so obtained are