If $(a, α)$ lies inside the circle $x^{2} + y^{2} = 9; x^{2} - 4 x - a^{2} = 0$ has exactly one root in $(- 1, 0)$ , then find the area of the region in which $(a, α)$ lies.

Open in App

Solution

If $(a, α)$ lies inside the circle
$x^{2} + y^{2} = 9$ therefore
$a^{2} + α^{2} < 9$ .......(1)
given $x^{2} - 4 x - a^{2} = 0$ has exactly one root in $(- 1, 0)$
$f (- 1) f (0) < 0$
$- (5 - a^{2}) a^{2} < 0$
$(a^{2} - 5) a^{2} < 0$ $a^{2}$ always greater than zero
$a^{2} < 5$ ......(2)
therefore $α^{2} < 4$
$| a | < \sqrt{5} | α | < 2$
$- \sqrt{5} < a < \sqrt{5} - 2 < α < 2$
Area of the region in which $(a, α)$ lies is
$= 4 \times 2 \sqrt{5}$
$= 8 \sqrt{5}$

Suggest Corrections

Similar questions

\frac{x}{α} + \frac{y}{β} = 1

x^{2} + y^{2} = a^{2},

(\frac{1}{α}, \frac{1}{β})

α

(α, 1)

C_{1} : x^{2} + y^{2} - 4 = 0

C_{2} : y^{2} = 4 x

(a, 0)

x^{2} + y^{2} = 4

x^{2} - 4 x - a^{2} = 0

(λ, - λ)

x^{2} + y^{2} - 4 x + 2 y - 8 = 0

λ

P (α, - α)

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1,

Join BYJU'S Learning Program