JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 12)
Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point ($$-$$5, 0). If the locus of the point P is a circle of radius r, then 4r2 is equal to ________
Answer
56
Explanation
Let P(h, k)
Given
PA = 3PB
PA2 = 9PB2
$$ \Rightarrow $$ (h $$-$$ 5)2 + k2 = 9[(h + 5)2 + k2]
$$ \Rightarrow $$ 8h2 + 8k2 + 100h + 200 = 0
$$ \therefore $$ Locus
$${x^2} + {y^2} + \left( {{{25} \over 2}} \right)x + 25 = 0$$
$$ \therefore $$ $$c \equiv \left( {{{ - 25} \over 4},0} \right)$$
$$ \therefore $$ $${r^2} = {\left( {{{ - 25} \over 4}} \right)^2} - 25$$
$$ = {{625} \over {16}} - 25$$
$$ = {{225} \over {16}}$$
$$ \therefore $$ $$4{r^2} = 4 \times {{225} \over {16}} = {{225} \over 4} = 56.25$$
After Round of 4r2 = 56
Given
PA = 3PB
PA2 = 9PB2
$$ \Rightarrow $$ (h $$-$$ 5)2 + k2 = 9[(h + 5)2 + k2]
$$ \Rightarrow $$ 8h2 + 8k2 + 100h + 200 = 0
$$ \therefore $$ Locus
$${x^2} + {y^2} + \left( {{{25} \over 2}} \right)x + 25 = 0$$
$$ \therefore $$ $$c \equiv \left( {{{ - 25} \over 4},0} \right)$$
$$ \therefore $$ $${r^2} = {\left( {{{ - 25} \over 4}} \right)^2} - 25$$
$$ = {{625} \over {16}} - 25$$
$$ = {{225} \over {16}}$$
$$ \therefore $$ $$4{r^2} = 4 \times {{225} \over {16}} = {{225} \over 4} = 56.25$$
After Round of 4r2 = 56
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