Question

Determiner as to what point the axes of the coordinates be shifted so as to remove the first degree terms from the equation

           (x, y) = 2x2 + 3y2 – 12x + 12+ 24 = 0

Solution

Correct option is

 

(3, –2)

Let the origin be shifted to (h, k).

Old New Pt.

       x = X + h

       y = Y + k

2(X + h)2 + 3(Y + k)2 – 12(X +h) + 12 (Y + k) + 24 = 0

2X2 + 3Y2 + X(2h – 12) + Y(6k + 12) + (12h2 + 3k2 – 12h + 12k) = 0

Since the equation is to be free of first degree terms, therefore coefficients of X and Y is each separately zero.

∴ 4h – 12 = 0 and 6k + 12 = 0

∴    (h, k) = (3, –2)

SIMILAR QUESTIONS

Q1

The value of p for which  is obtuse and  is a right angle is

Q2

A(p, 0), B(4, 0), C(5, 6) and D(1, 4) are the vertices of a quadrilateral ABCD 

If two sides of the quadrilateral are equal; area of the quadrilateral is

Q3

(0, 0), A(1, 1), B(0, 3) are the vertices of a triangle OAB.divides OB in the ratio 1 : 2, Q is the mid-point of AP, R divides AB in the ratio 2 : 1

1:- If 

Q4

(0, 0), A(1, 1), B(0, 3) are the vertices of a triangle OAB.divides OB in the ratio 1 : 2, Q is the mid-point of AP, R divides AB in the ratio 2 : 1

area of ΔPQR : area of ΔOAB is

Q5

(0, 0), A(1, 1), B(0, 3) are the vertices of a triangle OAB.divides OB in the ratio 1 : 2, Q is the mid-point of AP, R divides AB in the ratio 2 : 1

If S is the mid-point of PR, then QS is equal to

Q6

a and b real numbers between 0 and 1 A(a, 1), B(1, b) andC(0, 0) are the vertices of triangle

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Q7

If the triangle ABC in  isosceles with AC = BC and 5(AB)2 = 2(AC)2 then

Q8

The origin is shifted to(1, –2)then what are the coordinates be shifted if the point (3, –5) in the new position?

Q9

If the origin is shifted to (1, –2), the coordinates of A become (2, 3). What are the original coordinates of A?

Q10

What will be the coordinates of the point  when the axes are rotated through an angle of 300 in clockwise sense?