Find the value of k, if the line joining the origin to the points of intersection of the curve 2 2 2 2 3 2 1 0 x xy y x y and the line x y k 2 are mutually perpendicular.
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Answer:
Given that,
the curve
2x
2
−2xy+3y
2
+2x−y−1=0 __(1)
the line
x+2y=k
k
x+2y
=1 __(2)
By equation (1) to
(2x
2
−2xy+3y
2
)+(2x−y)1−1(1)
2
From equation(2) to,
(2x
2
−2xy+3y
2
)+(2x−y)
k
(x+2y)
−1(
k
x+2y
)
2
=0
k
2
(2x
2
−2xy+3y
2
)+k(2x−y)(x+2y)−(x+2y)
2
=0
2k
2
x
2
−2k
2
xy+3k
2
y
2
+2kx
2
+3kxy−2ky
2
−x
2
−4xy−4y
2
=0
⇒x
2
(2k
2
+2k−1)+xy(−2k
2
+3k−4)+y
2
(3k
2
−2k−4)=0
These lines are given mutually perpendicular
∴x
2
coeff.+y
2
coeff=0
⇒2k
2
+2k−1+3k
2
−2k−4=0
⇒5k
2
−5=0
k
2
−1=0 , k=±1
Step-by-step explanation:
hope i helped!(PS: don't know why it wrote it like this)
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