the point on the hyperbola  x² / 24  -  y² / 18 = 1  which is nearest to the line  3x +2y + 1 = 0 is

cant get the question man!

there will be some point which is on the line as well as on the hyperbola and can be obtained by solving the 2 eqns. diatance of that pt will be 0 from the hyperbola.

intimate me if i m correct.

Is the ans (6,-3).If correct,I will explain.

The general point on Hyperbola is (2rt6seck,3rt2tank).Calculate the dis. of this point from the line using the formula and it proves to be min iff rt3seck+tank is min.

The critical points occur at sink=-1/rt3 and there are two points one in 2nd and other in 4th quadrant[plot a rough graph] where the distance might be min.If we substitute corresponding values of seck and tank taking the corresponding signs,the dis comes out to be min for the point in Q₄  and the point is (6,-3) 

Distance=I6rt6seck+6rt2tank+1I/rt13=1/rt13[6rt2(rt3seck+tank)+1].So,the distance is min iff rt3seck+tank is min as remaining terms are constants.

Parametric points of the given hyperbola are (root(24)sec , root(18)tan

distance of this point from the given line is



(3root24sec + 2root18tan + 1 ) / root(13)....let this be f()



so for critical points f'()=0

so f'() = 3root(24/13)tansec + 2root(18/13)sec^2 = 0

((3root24)sin + 2root18) / (root13(cos^2)) = 0



sin = -2root18/3root24 = -1/root3

so point is  (root(24)sec , root(18)tan)  (6,-3)