Question

(b) Find the value of k such that point (3,-2) lies on the locus 2x²-3y² + 2k-6y + 2 = 0

Answer 1

⇒ k=−9

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Explanation:

then put P(−2, 2) and we get,

x

2

−7x+ky=0

(−2)

2

−7(−2)+k(2)=0

⇒ 4+14+2k=0

⇒ 18+2k=0

⇒ 2k=−18

⇒ k=−9

again point Q(3, a) lies on the locus

the, put Q(3, a) and k=−9. We get

x

2

−7x+ky=0

(3)

2

−7(3)+(−9)×a=0

9−21−9a=0

−9a=+12

a=

9

−12

​

a=

3

−4

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Answer 2

`{ \tt{ {2x}^(2) - {3y}^(2) + 2k - 6y + 2 = 0 }} \\ { \tt{ \{x = 3 \: \: and \: \: y = - 2 \}}} \\ { \tt{ {2(3)}^(2) - {3( - 2)}^(2) + 2k - 6( - 2) + 2 = 0}} \\ { \tt{18 - 12 + 2k + 12 + 2 = 0}} \\ { \tt{2k + 20 = 0}} \\ { \tt{2k = - 20}} \\ { \tt{k = - 10}}`