$Find the point to which the origin should be dhifted after a translation of axes so that the following equations will have no first degree terms: ? (i) y^{2} + x^{2} - 4 x - 8 y + 3 = 0 (i i) x^{2} + y^{2} - 5 x + 2 y - 5 = 0 (i i i) x^{2} - 12 x + 4 = 0$

Open in App

Solution

$(i) Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k Substituting x = X + h, y = Y + k in the equation, y^{2} + x^{2} - 4 x - 8 y + 3 = 0, we get (Y + k)^{2} + (X + h)^{2} - 4 (X + h) - 8 (Y + k) + 3 = 0 \Rightarrow Y^{2} + k^{2} + 2 Y k + X^{2} + h^{2} + 2 X h - 4 X - 4 h - 8 Y - 8 k + 3 = 0 \Rightarrow Y^{2} + X^{2} + 2 Y k - 8 Y + 2 X h - 4 X + k^{2} + h^{2} - 4 h - 8 k + 3 = 0 \Rightarrow Y^{2} + X^{2} + (2 k - 8) Y + (2 h - 4) X + (k^{2} + h^{2} - 4 h - 8 k + 3) = 0 For this equation to be free from the term of first degree, we must have 2k - 8 = 0 and 2h - 4 = 0 \Rightarrow k = 4 a n d h = 2 Hence, the origin is shifted at the point (2, 4.) (i i) x^{2} + y^{2} - 5 x + 2 y - 5 = 0 Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k Substituting x = X + h, y = Y + k in the equation, x^{2} + y^{2} - 5 x + 2 y - 5 = 0, we get (X + h)^{2} + (Y + k)^{2} - 5 (X + h) + 2 (Y + k) - 5 = 0 \Rightarrow X^{2} + h^{2} + 2 X h + Y^{2} + k^{2} + 2 Y k - 5 X - 5 h + 2 Y + 2 k - 5 = 0 \Rightarrow X^{2} + Y^{2} + 2 Y k + 2 Y + 2 X h - 5 X + h^{2} + k^{2} - 5 h + 2 k - 5 = 0 \Rightarrow X^{2} + Y^{2} + (2 k + 2) Y + (2 h - 5) X + h^{2} + k^{2} - 5 h + 2 k - 5 = 0 For this equation to be free from the term of first degree, we must have 2k + 2 = 0 and 2h - 5 = 0 \Rightarrow k = - 1 a n d h = \frac{5}{2} Hence, the origin is shifted at the point (\frac{5}{2}, - 1) . (i i i) x^{2} - 12 x + 4 = 0 Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k Substituting x = X + h, y = Y + k in the equation, x^{2} + 12 x + 4 = 0, we get (X + h)^{2} - 12 (X + h) + 4 = 0 \Rightarrow X^{2} + h^{2} + 2 X h - 12 X - 12 h + 4 = 0 \Rightarrow X^{2} + (2 h - 12) X + h^{2} - 12 h + 4 = 0 For this equation to be free from the term of first degree, we must have 2h - 12 = 0 \Rightarrow h = \frac{12}{2} \Rightarrow h = 6 Hence, the origin is shifted at the point (6, k) K \in R .$

Suggest Corrections

Similar questions

Q. Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first degree terms:
(i) y² + x² − 4x − 8y + 3 = 0
(ii) x² + y² − 5x + 2y − 5 = 0
(iii) x² − 12x + 4 = 0

Find the point to which the origin be shifted after a translation, so that the equation $x^{2} + y^{2} - 4 x - 8 y + 3 = 0$ will have no first degree terms.

Q. Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:
(i)

x^{2} + 2 y^{2} - 2 x + 12 y + 10 = 0

(ii)

x^{2} + 4 y^{2} - 4 x + 24 y + 31 = 0

(iii)

4 x^{2} + y^{2} - 8 x + 2 y + 1 = 0

(iv)

3 x^{2} + 4 y^{2} - 12 x - 8 y + 4 = 0

(v)

4 x^{2} + 16 y^{2} - 24 x - 32 y - 12 = 0

(vi)

x^{2} + 4 y^{2} - 2 x = 0

Find the equation of radical axis of two circle $x^{2} + y^{2} - 4 x - 2 y = 4$ and $x^{2} + y^{2} - 12 x - 8 y = 12.$

Q. The number of common tangents that can be drawn to the circles

x^{2} + y^{2} - 12 x + 8 y + 48 = 0

and

x^{2} + y^{2} - 4 x + 2 y - 4 = 0

Join BYJU'S Learning Program

Find the point to which the origin should be dhifted after a translation of axes so that the following equations will have no first degree terms:?(i) y2+x2−4x−8y+3=0(ii) x2+y2−5x+2y−5=0(iii) x2−12x+4=0

$Find the point to which the origin should be dhifted after a translation of axes so that the following equations will have no first degree terms: ? (i) y^{2} + x^{2} - 4 x - 8 y + 3 = 0 (i i) x^{2} + y^{2} - 5 x + 2 y - 5 = 0 (i i i) x^{2} - 12 x + 4 = 0$