First slide

Hyperbola in conic sections

Question

A hyperbola $H : \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ ntersects the circle, $C : x^{2} + y^{2} - 8 x = 0$ at the points A and B.
Statement-1: $2 x - \sqrt{5} y + 4 = 0$ is a common tangentto both C and H.
Statement-2: Circle on AB as a diameter passes through the centre of the hyperbola H.

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Equation of a tangent with slope $\frac{2}{\sqrt{5}}$ $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ is $y = \frac{2}{\sqrt{5}} x + \sqrt{9 \times \frac{4}{5} - 4} \Rightarrow 2 x - \sqrt{5} y + 4 = 0$
Next $2 x - \sqrt{5} y + 4 = 0$ touches the circle $(x - 4)^{2} + y^{2}$ = 16 if the length of the perpendicular from (4, 0) on the line is 4 which is true. Hence statement-1 is true. In statement-2, let A be $(3 \sec θ, 2 \tan θ)$
A lies on the circle $x^{2} + y^{2} - 8 x = 0$
$\Rightarrow 13 \sec^{2} θ - 24 \sec θ - 4 = 0$
$\Rightarrow \sec θ = 2 \Rightarrow \tan θ = \pm \sqrt{3}$
So the coordinate of A are $(6, 2 \sqrt{3})$ and of $B$ are
$(6, - 2 \sqrt{3})$ and equation of the circle on AB as diameter is $(x - 6) (x - 6) + (y - 2 \sqrt{3}) (y + 2 \sqrt{3}) = 0$
$\Rightarrow x^{2} + y^{2} - 12 x + 24 = 0$ which does not pass through the centre (0, 0) of the hyperbola. Thus statement-2 is false.

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