JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 3)
The locus of the centroid of the triangle formed by any point P on the hyperbola $$16{x^2} - 9{y^2} + 32x + 36y - 164 = 0$$, and its foci is :
$$16{x^2} - 9{y^2} + 32x + 36y - 36 = 0$$
$$9{x^2} - 16{y^2} + 36x + 32y - 144 = 0$$
$$16{x^2} - 9{y^2} + 32x + 36y - 144 = 0$$
$$9{x^2} - 16{y^2} + 36x + 32y - 36 = 0$$
Explanation
Given hyperbola is
$$16{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144$$
$$ \Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1$$
Eccentricity, $$e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}$$
$$\Rightarrow$$ foci are (4, 2) and ($$-$$6, 2)
_25th_July_Morning_Shift_en_3_1.png)
Let the centroid be (h, k) & A($$\alpha$$, $$\beta$$) be point on hyperbola.
So, $$h = {{\alpha - 6 + 4} \over 3},k = {{\beta + 2 + 2} \over 3}$$
$$ \Rightarrow \alpha = 3h + 2,\beta = 3k - 4$$
($$\alpha$$, $$\beta$$) lies on hyperbola so
$$16{(3h + 2 + 1)^2} - 9{(3k - 4 - 2)^2} = 144$$
$$ \Rightarrow 144{(h + 1)^2} - 81{(k - 2)^2} = 144$$
$$ \Rightarrow 16({h^2} + 2h + 1) - 9({k^2} - 4k + 4) = 16$$
$$ \Rightarrow 16{x^2} - 9{y^2} + 32x + 36y - 36 = 0$$
$$16{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144$$
$$ \Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1$$
Eccentricity, $$e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}$$
$$\Rightarrow$$ foci are (4, 2) and ($$-$$6, 2)
Let the centroid be (h, k) & A($$\alpha$$, $$\beta$$) be point on hyperbola.
So, $$h = {{\alpha - 6 + 4} \over 3},k = {{\beta + 2 + 2} \over 3}$$
$$ \Rightarrow \alpha = 3h + 2,\beta = 3k - 4$$
($$\alpha$$, $$\beta$$) lies on hyperbola so
$$16{(3h + 2 + 1)^2} - 9{(3k - 4 - 2)^2} = 144$$
$$ \Rightarrow 144{(h + 1)^2} - 81{(k - 2)^2} = 144$$
$$ \Rightarrow 16({h^2} + 2h + 1) - 9({k^2} - 4k + 4) = 16$$
$$ \Rightarrow 16{x^2} - 9{y^2} + 32x + 36y - 36 = 0$$
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