Match the conics in Column I with the statements/expressions in Column II. The locus of the point for which the line touches the circle [2009]

Question

Match the conics in Column I with the statements/expressions in Column II. [2009]

	Column I		Column II
(A)	Circle	(p)	The locus of the point $(h, k)$ for which the line $h x + k y = 1$ touches the circle $x^{2} + y^{2} = 4$
(B)	Parabola	(q)	Points $z$ in the complex plane satisfying $\| z + 2 \|$ $-$ $\| z - 2 \|$ $= \pm 3$
(C)	Ellipse	(r)	Points of the conic have parametric representation $x = \sqrt{3} (\frac{1 - t^{2}}{1 + t^{2}}), y = \frac{2 t}{1 + t^{2}}$
(D)	Hyperbola	(s)	The eccentricity of the conic lies in the interval $1 \leq x < \infty$
		(t)	Points $z$ in the complex plane satisfying $Re {(z + 1)}^{2} = \| z \|^{2} + 1$

Smart Achievers · Accepted Answer

(1)

$(p) As the line h x + k y = 1, touches the circle x^{2} + y^{2} = 4$

$∴$ $Length of perpendicular from centre (0, 0) of circle to the line$ $= radius of the circle$

$\Rightarrow \frac{1}{\sqrt{h^{2} + k^{2}}} = 2$ $\Rightarrow h^{2} + k^{2} = \frac{1}{4}$

$∴$ $Locus of (h, k) is x^{2} + y^{2} = \frac{1}{4}, which is a circle.$

$(q) We know that if | z - z_{1} | - | z - z_{2} | = k,$

where $| k | < | z_{1} - z_{2} |,$ then $z$ traces a hyperbola.

Here $| z + 2 | - | z - 2 | = \pm 3$

$∴$ $Locus of z is a hyperbola.$

$(r) Given : x = \sqrt{3} (\frac{1 - t^{2}}{1 + t^{2}}), y = \frac{2 t}{1 + t^{2}}$

$\Rightarrow \frac{x}{\sqrt{3}} = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$

On squaring and adding, we get

$\frac{x^{2}}{3} + y^{2} = \frac{{(1 - t^{2})}^{2} + 4 t^{2}}{{(1 + t^{2})}^{2}} = 1$ $\Rightarrow \frac{x^{2}}{3} + \frac{y^{2}}{1} = 1$

which is the equation of an ellipse.

$(s) We know, eccentricity of a parabola = 1$

$and eccentricity of an ellipse < 1$

$and eccentricity of a hyperbola > 1 .$

$Hence, the conics whose eccentricity lies in$ $1 \leq x < \infty$ $are parabola and hyperbola.$

$(t) Let z = x + i y$

$∴$ $Re {[(x + 1) + i y]}^{2} = x^{2} + y^{2} + 1$

$\Rightarrow {(x + 1)}^{2} - y^{2} = x^{2} + y^{2} + 1$

$∴$ $y^{2} = x,$ which is a parabola.

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