KVPY SA Exam 2018 Question Paper

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Question : 10 of 80

Marks: +1, -0

Let

x_0, y_0

be fixed real numbers such that

x _0^2+y_0^2 > 1

. If x,y are arbitrary real numbers such that

x^2 + y^2 ≤ 1,

then the minimum value of

(x -x_0)^ 2 + (y -y_0)^ 2

is

$(√{x _0^2+y_0^2} - 1)^2$
${x _0^2+y_0^2} - 1$
$({|x _0| +|y_0 |} - 1) ^2$
$({|x _0| +|y_0 |} )^2- 1$

Solution:

x _0^2+y_0^2 > 1\;\;\;\;\;x _0 -y _0 \fixed

x, y \;\;\;\;\;\arbitrary

x _0^2+y_0^2 ≤ 1\;\;\; ,\Let\;\;\;\;x = \cosθ,

\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\y = \sinθ

for men

z = (x -x_0)^ 2 + (y -y_0)^ 2

z = x^2 + x_0^ 2 + y^2 + y_0^ 2 -2 (x\ x_0 + y\ y_0)

put x - cosθ , y = sinθ

z = x _ 0^ 2+ y _0^2-2 (x_0 \cosθ + y_0 \sinθ)

{dz}/{dθ}⇒0-2 (-x _0 \sinθ + y _0 \cosθ)

{dz}/{dθ}=0

-x_0 \sinθ = -y_0 \cosθ

\tanθ=y_0/x_0

\sinθ=y_0/{√{x_0^2+y_0^2}

\cos θ=x_0/{√{x_0^2+y_0^2}

x= {x_0}/{√{x_0^2+y_0^2}},y=y_0 /{√{x_0^2+y_0^2}

z= (x_0/ √{x_0^2+y_0^2}-x_0)^2

+(y_0/ √{x_0^2+y_0^2}-y_0)^2

= x_0 2(x_0/ √{x_0^2+y_0^2}-1)^2

+y_0 2(y_0/ √{x_0^2+y_0^2}-1)^2

=({x_0^2+y_0^2})(1-√{x_0^2+y_0^2})/(√{x_0^2+y_0^2})^2

(1-√{x_0^2+y_0^2})⇒( √{x_0^2+y_0^2}-1)^2

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