JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 8)
Let O(0, 0) and A(0, 1) be two fixed points. Then
the locus of a point P such that the perimeter of
$$\Delta $$AOP is 4, is :
9x2 + 8y2 – 8y = 16
8x2 – 9y2 + 9y = 18
8x2 + 9y2 – 9y = 18
9x2 – 8y2 + 8y = 16
Explanation
_8th_April_Morning_Slot_en_8_2.png)
Let point C(h, k)
Given, AB + BC + AC = 4
From graph, AB = 1
$$ \therefore $$ BC + AC = 3
$$ \Rightarrow $$ $$\sqrt {{h^2} + {k^2}} + \sqrt {{h^2} + {{\left( {k - 1} \right)}^2}} = 3$$
$$ \Rightarrow $$ $${{h^2} + {{\left( {k - 1} \right)}^2}}$$ = 9 + $${{h^2} + {k^2}}$$ - $$6\sqrt {{h^2} + {k^2}} $$
$$ \Rightarrow $$ $$6\sqrt {{h^2} + {k^2}} = 2k + 8$$
$$ \Rightarrow $$ $$9\left( {{h^2} + {k^2}} \right) = {k^2} + 8k + 16$$
$$ \Rightarrow $$ $$9{h^2} + 8{k^2} - 8k - 16 = 0$$
$$ \therefore $$ Locus of point C will be,
9x2 + 8y2 – 8y = 16
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