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Focii of Ellipse

A variable li...

Question

A variable line cuts positive $X - a x i s$ and $Y - a x i s$ at $A$ and $B$ respectively such that $3 (O A) + 4 (O B) + O A . O B = 10$ then the locus of foot of perpendicular drawn from $(0, 0)$ onto this line is

(x^{2} + y^{2}) (x + y + 7) = 8 x y

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(x + y) (3 x + 4 y) = x y

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(x^{2} + y^{2}) (x^{2} + y^{2} + 3 x + 4 y) = 10 x y

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(x^{2} + y^{2}) (x^{2} + y^{2} + 4 x + 3 y) = 10 x y

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Solution

The correct option is D $(x^{2} + y^{2}) (x^{2} + y^{2} + 4 x + 3 y) = 10 x y$

From the question, we can get

$\frac{x}{h} + \frac{y}{k} = 1$

$3 h + 4 k + h k = 10$

Equation of line:

$y - β = - \frac{α}{β} (x - α)$

$β y - β^{2} = - α x + α^{2}$

$α x + β y = α^{2} + β^{2}$

$\frac{x}{(\frac{α^{2} + β^{2}}{α})} + \frac{y}{(\frac{α^{2} + β^{2}}{β})} = 1$

$h = \frac{α^{2} + β^{2}}{α} k = \frac{α^{2} + β^{2}}{β}$

$3 \frac{α^{2} + β^{2}}{β} + 4 \frac{α^{2} + β^{2}}{β} + \frac{{(α^{2} + β^{2})}^{2}}{α β} = 10$

$3 β (α^{2} + β^{2}) + 4 α (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = 10 α β$

$(α^{2} + β^{2}) [4 α + 3 β + α^{2} + β^{2}] 10 α β$

$(x^{2} + y^{2}) [4 x + 3 y + x^{2} + y^{2}] = 10 x y$

$H e n c e, o p t i o n D i s t h e c o r r e c t a n s w e r .$

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