If the coordinates of a point are where is parameter then the points lies on a conic whose eccentricity is

Question

Match the items of Column-I with Column-II

	Column-I		Column-II
(I)	If the coordinates of a point are $(4 \tan θ, 3 \sec θ)$ where $θ (θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$ is parameter then the points lies on a conic whose eccentricity is	P	$\sqrt{3}$
(II)	If an ellipse has the length of major axis 10 units and semi minor axis length 4 units, then this ellipse has eccentricity	Q	$\frac{4}{5}$
(III)	If AB is double ordinate of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{9} = 1$ such that triangle OAB is an equilateral triangle of side '2' then eccentricity of hyperbola is (where O is centre of Hyperbola)	R	$\frac{5}{3}$
(IV)	If the foci of ellipse $\frac{x^{2}}{K^{2} a^{2}} + \frac{y^{2}}{a^{2}} = 1$ and hyperbol $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1$ coincide then K can be	S	$\frac{3}{5}$
		T	$\sqrt{\frac{13}{3}}$

Smart Achievers · Accepted Answer

(1)

$I)$ $\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$ $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ $e = \sqrt{\frac{16 + 9}{16}} = \frac{5}{4}$ $\frac{1}{e^{2}} + \frac{1}{e'^{2}} = 1$

$\frac{1}{e'^{2}} = 1 - \frac{16}{25} = \frac{9}{25}$ $e' = \frac{5}{3}$

$II)$ $2 a = 10 \Rightarrow a = 5$ $b^{2} = 16, b = 4$ $e = \sqrt{\frac{25 - 16}{25}}$

$III)$ $(a \sec θ, 3 \tan θ)$

$6 \tan θ = 2$ $\tan θ = \frac{1}{3}$ $\frac{1}{\sqrt{3}} = \frac{3 \tan θ}{a \sec θ}$

$a^{2} \sec^{2} θ = 27 \cdot \frac{1}{9}$

$a^{2} (1 + \frac{1}{9}) = 27 \cdot \frac{1}{9}$ $a^{2} (10) = 27$

$IV)$ $k^{2} a^{2} - a^{2} = a^{2} + a^{2}$

$k^{2} - 1 = 2 \Rightarrow k^{2} = 3 \Rightarrow \sqrt{3}$

Solution

1 I-R; II-S; III-T; IV-P

2 I-S; II-P; III-Q; IV-T

3 I-P; II-T; III-S; IV-Q

4 I-Q; II-S; III-P; IV-T

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