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The line 3x+6y=k intersects the curve 2x...

The line `3x+6y=k` intersects the curve `2x^2+3y^2=1` at points `Aa n dB` . The circle on `A B` as diameter passes through the origin. Then the value of `k^2` is__________

A

3

B

4

C

`-4`

D

`-3`

Text Solution

Verified by Experts

The correct Answer is:
A, D

We have `(3x +6y)/(k) = 1` (1)
`2x^(2) +2xy +3y^(2) -1 = 0` (2)
Now homogenising (2) with the help of (1), we get
`rArr 2x^(2) +2xy +3y^(2) -((3x +6y)/(k))^(2) =0`
`rArr k^(2)(2x^(2)+3x +3y^(2)) -(3x +6y)^(2) =0`
Now, coefficient of `x^(2)+` coefficient of `y^(2) =0`
`rArr (2k^(2)-9) +(3k^(2) -36) =0`
`rArr 5k^(2) =45 rArr k^(2) = 9 rArr k = 3` or `-3`
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Knowledge Check

  • The point of intersection of the curves y^(2) = 4x and the line y = x is

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