Minimum distance...
{ x 2 + y 2 + 6 x − 1 2 y + 2 0 = 0 y 2 = 4 x
The shortest distance between the above curves can be written in the form a b − c , then find a + b + c .
Note: a , b and c are positive integers and b is not a perfect square.
The answer is 11.
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2 solutions
Why is it a level 5 problem ?
Nice Solution :)
Nice solution.
Why is the slope of the common normal -a ?
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Sorry dude my mistake and i corrected it. Its 2 − α .
The slope of line tangent at point ( − a 2 , a ) on parabola will be d x d y at that point.
= > ( 2 ) ( y ) ( d x d y ) = 4
= > d x d y = y 2
=> Now the y coordinate is a so slope of tangent in parabola will be a 2 and this is perpendicular to the normal so the slope of normal will be 2 − a .
=> By mistake yash would have written the slope as -a:)
y=mx-2am-am^3 is general eqn. for normal of parabola given.this should pass through centre of circle i.e.(-3,6).from this we get slope of normal as -1.therefore slope os tangent at that point is 1 =2/y using application of derivative ,now thus x is 1.u got points now .thus find its distance from centre and substract radius of circle to get answer =4sqrt2-5
Let the co-ordinates of the point P be ( 4 α 2 , α ) . Now, the shortest distance between both the curves will be the common normal. The slope of common normal will be 2 − α .
=>\quad \frac { -\alpha }{ 2 } =\frac { \alpha -6 }{ { \frac { { \alpha }^{ 2 } }{ 4 } }+3 } \\ =>\quad -{ \alpha }^{ 3 }-12\alpha =8\alpha -48\\ =>{ \quad \alpha }^{ 3 }+20\alpha -48=0\\ =>\quad \alpha =2\\
Now the co-ordinates of P are ( 1 , 2 ) . Minimum distance will be ( 1 − ( − 3 ) ) 2 + ( 2 − 6 ) 2 − 5 = 1 6 + 1 6 − 5 = 4 2 − 5 . Hence answer is 4 + 2 + 5 = 1 1
Please don't mind the diagram...:P