6.
If the straight lines joining the origin and the points of intersection of the curve 5x + 12xy-6y + 4x-2y+3=0
and x + ky - 1 = 0 are equally inclined to the co-ordinate axes then the value of k :
(A) is equal to
(B) is equal to - 1
(C) is equal to 2
(D) does not exist in the set of real numbers.
2
2
Answers
Answer:
5x
2
+12xy−6y
2
+4x−2y+3=0 ------ ( 1 )
x+ky−1=0
x+ky=1 ----- ( 2 )
Hamoginizing ( 1 ) with ( 2 )
⇒ 5x
2
+12xy−6y
2
+4x(x+ky)−2y(x+ky)+3(x+ky)
2
=0
⇒ 5x
2
+12xy−6y
2
+4x
2
+4kxy−2xy−2ky
2
+3(x
2
+kxy+k
2
y
2
)=0
⇒ 9x
2
+10xy−6y
2
+4kxy−2ky
2
+3x
2
+3kxy+3k
2
y
2
=0
⇒ 12x
2
+(10+4k+6k)xy+(3k
2
−2k−6)y
2
=0
⇒ 12x
2
+(10+10k)xy+(3k
2
−2k−6)y
2
=0
Pair of lines are equally inclined to the axes.
∴ Coefficient of xy=0
⇒ 10k+10=0
⇒ 10k=−10
⇒ k=−1
∴ ∣k∣=∣−1∣=1
Step-by-step explanation:
1
Answer:
I think that you meant (12xy - 6y^2). That's because it's the general question. Sorry that I don't know how to solve it myself but I hope that this information will help you.
Step-by-step explanation: