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Find the point on the curve 3x^2-4y^2=72...

Find the point on the curve `3x^2-4y^2=72` which is nearest to the line `3x+2y+1=0.`

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Find the points on the curve 3x^(2)-4y^(2)=72 which is nearest t the line 3x+2y+1=0.

If M(x_(0), y_(0)) is the point on the curve 3x^(2)-4y^(2)=72 which is nearest to the line 3x+2y+1=0 then the value of (x_(0)+y_(0)) is equal to

If M(x_(o),y_(o)) is the point on the curve 3x^(2)-4y^(2)=72 which is nearest to the line 3x+2y+1=0 ,then the value of (x_(o)+y_(o)) is equal to (A)3(B)-3(C)9(D)-9

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