106x2=2y
Any point can be taken as x=2t y=2t2
distance of any point from (0,3) = √4t2+(2t2-3)2
= √4t4-8t2+9
The min value of this will be attained at t2=-b/a = 12/8 = 3/2.
So t=±√1.5
So the point is (±√6,3)
1.the point (0,3) is nearest to the curve x^2=2y at
2.in a triangle ABC ANGLE B=90 and a+b=4 .te area of the triangle is the maximum when angle C is
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2 Answers
106