All chords of the curve \( 3 x^{2}-y^{2}+6 x+2 y=0 \) which subtend a

1.	All chords of the curve \( 3 x^{2}-y^{2}+6 x+2 y=0 \) which subtend a right angle at the origin always passes through a fixed point \( (\alpha, \beta) \)A) \( \|\alpha\|=\|\beta\| \)B) \( \|\alpha\|>\|\beta\| \)C) \( \|\alpha\|
Answer» Given curve is 3x² - y³ + 6x + 2y = 0.....(i) Let the chord be y = mx + c ⇒ \(\frac{y + mx}c=1\) Given that chord are substanded a right angle at the origin. By using homogenisation Since, 3x² - y² + (3x + y) X 1 = 0 ⇒ 3x² - y² + 2(3x + y)\((\frac{y + mx}c)\) = 0 \((\because \frac{y-mx}c=1)\) ⇒ 3x²c - cy² + 6xy - 6mx² + 2y² - 2mxy = 0 ⇒ (3c - 6m)x² + (6 - 2m)xy + (2 - c)y² = 0......(ii) Since, Curve (ii) is subtended a right angle at the origin. Therefore, Coefficient of x² + coefficient of y² = 0 ⇒ 3c - 6m + 2 - c = 0 ⇒ 2c - 6m + 2 = 0 ⇒ 2 = 6m - 2c........(iii) Given chord is y = mx + c Therefore for fixed point, \(\frac{2}y=\frac6x=\frac{-2}1\) ⇒ x = \(\frac{-6}2\) = -3 and \(\frac2y\) = -2 ⇒ y = \(\frac2{-2}\) = -1 Hence, fixed point is (-3, -1) ≡ (α, β) (given) ⇒ α = -3, β = -1 ⇒ \|α\| = 3, \|β\| = 1 ⇒ \|α\| > \|β\| option (B) is correct.

All chords of the curve \( 3 x^{2}-y^{2}+6 x+2 y=0 \) which subtend a right angle at the origin always passes through a fixed point \( (\alpha, \beta) \)A) \( |\alpha|=|\beta| \)B) \( |\alpha|>|\beta| \)C) \( |\alpha|

Answer»

Given curve is 3x² - y³ + 6x + 2y = 0.....(i)

Let the chord be y = mx + c

⇒ \(\frac{y + mx}c=1\)

Given that chord are substanded a right angle at the origin.

By using homogenisation

Since, 3x² - y² + (3x + y) X 1 = 0

⇒ 3x² - y² + 2(3x + y)\((\frac{y + mx}c)\) = 0

\((\because \frac{y-mx}c=1)\)

⇒ 3x²c - cy² + 6xy - 6mx² + 2y² - 2mxy = 0

⇒ (3c - 6m)x² + (6 - 2m)xy + (2 - c)y² = 0......(ii)

Since, Curve (ii) is subtended a right angle at the origin.

Therefore, Coefficient of x² + coefficient of y² = 0

⇒ 3c - 6m + 2 - c = 0

⇒ 2c - 6m + 2 = 0

⇒ 2 = 6m - 2c........(iii)

Given chord is y = mx + c

Therefore for fixed point,

\(\frac{2}y=\frac6x=\frac{-2}1\)

⇒ x = \(\frac{-6}2\) = -3

and \(\frac2y\) = -2

⇒ y = \(\frac2{-2}\) = -1

Hence, fixed point is (-3, -1) ≡ (α, β) (given)

⇒ α = -3, β = -1

⇒ |α| = 3, |β| = 1

⇒ |α| > |β|

option (B) is correct.