Question

27. The point on the curve x2- 2y which is nearest to the point (0, 5) is(A) (2V2,4) (B) (2^2,0) (C) (o.0) (D) (2,2)

Open in App
Solution

The given equation is given as, x 2 =2y(1) It is given that the curve is nearest to the point ( 0,5 ). Consider point P( x,y ) as any point on the curve. The formula for the distance between two points is, d= ( x− x 1 ) 2 + ( y− y 1 ) 2 Substitute 0 for x 1 and 5 for y 1 in the above equation. d= ( x−0 ) 2 + ( y−5 ) 2 d 2 = x 2 + ( y−5 ) 2 Substitute the value x 2 =2y from equation (1) d 2 =2y+ ( y−5 ) 2 d 2 =2y+ y 2 +25−10y d 2 = y 2 −8y+25 Consider, d 2 =M M= y 2 −8y+25 For maxima or minima, dM dy =0 dM dy =2y−8 2y−8=0 y=4 Also d 2 M d y 2 =2 =+ve Hence, M is minimum and d is minimum at y=4. From equation (1) x 2 =8 x=±2 2 Therefore, ( 2 2 ,4 ) and ( −2 2 ,4 ) are two points on curve nearest to (0,5). Thus, the correct option is (A).

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Second Derivative Test for Local Minimum
MATHEMATICS
Watch in App
Join BYJU'S Learning Program