JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 14)
If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is -
$$\sqrt 3 $$
$${{\sqrt 3 } \over 2}$$
$${2 \over {\sqrt 3 }}$$
$${1 \over {\sqrt 3 }}$$
Explanation
Area bounded by
y2 = 4ax & x2 = 4by, a, b $$ \ne $$ 0
is $$\left| {{{16ab} \over 3}} \right|$$
by using formula :
4a $$=$$ $${1 \over k} = 4b,k > 0$$
Area $$ = \left| {{{16.{1 \over {4k}}.{1 \over {4k}}} \over 3}} \right| = 1$$
$$ \Rightarrow $$ k2 $$ = {1 \over 3}$$
$$ \Rightarrow $$ k $$ = {1 \over {\sqrt 3 }}$$
y2 = 4ax & x2 = 4by, a, b $$ \ne $$ 0
is $$\left| {{{16ab} \over 3}} \right|$$
by using formula :
4a $$=$$ $${1 \over k} = 4b,k > 0$$
Area $$ = \left| {{{16.{1 \over {4k}}.{1 \over {4k}}} \over 3}} \right| = 1$$
$$ \Rightarrow $$ k2 $$ = {1 \over 3}$$
$$ \Rightarrow $$ k $$ = {1 \over {\sqrt 3 }}$$
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