JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 17)
Let $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, where a > b > 0, be a hyperbola in the XY-plane whose conjugate axis LM subtends an angle of 60$$^\circ $$ at one of its vertices N. Let the area of the $$\Delta $$LMN be $$4\sqrt 3 $$.
| List - I | List - II | ||
|---|---|---|---|
| P. | The length of the conjugate axis of H is | 1. | 8 |
| Q. | The eccentricity of H is | 2. | $${4 \over {\sqrt 3 }}$$ |
| R. | The distance between the foci of H is | 3. | $${2 \over {\sqrt 3 }}$$ |
| S. | The length of the latus rectum of H is | 4. | 4 |
P $$ \to $$ 4 ; Q $$ \to $$ 2 ; R $$ \to $$ 1 ; S $$ \to $$ 3
P $$ \to $$ 4 ; Q $$ \to $$ 3 ; R $$ \to $$ 1 ; S $$ \to $$ 2
P $$ \to $$ 4 ; Q $$ \to $$ 1 ; R $$ \to $$ 3 ; S $$ \to $$ 2
P $$ \to $$ 3 ; Q $$ \to $$ 4 ; R $$ \to $$ 2 ; S $$ \to $$ 1
Explanation
We have,
Equation of hyperbola
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
It is given,
$$\angle LNM = 60^\circ {4 \over {\sqrt 3 }}$$
and Area of $$\Delta LMN = 4\sqrt 3 $$
Now, $$\Delta LMN$$ is an equilateral triangle whose sides is 2b [$$ \because $$ $$\Delta LON$$ $$ \cong $$ $$\Delta MOL$$; $$ \because $$ $$\angle NLO = \angle NMO = 60^\circ $$]
$$ \therefore $$ Area of $$\Delta LMN = {{\sqrt 3 } \over 4}{(2b)^2}$$
$$ \Rightarrow $$ $$4\sqrt 3 = \sqrt 3 {b^2} \Rightarrow b = 2$$
Also, area of $$\Delta LMN = {1 \over 2}a(2b) = ab$$
$$ \Rightarrow $$ $$4\sqrt 3 = a(2)$$
$$ \Rightarrow $$ $$a = 2\sqrt 3 $$
(P) Length of conjugate axis = 2b = 2(2) = 4 (Q)
Eccentricity (e) = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {4 \over {12}}} $$
= $${4 \over {2\sqrt 3 }} = {2 \over {\sqrt 3 }}$$
(R) Distance between the foci
= $$2ae = 2 \times 2\sqrt 3 \times {2 \over {\sqrt 3 }} = 8$$
(S) The length of latusrectum
= $${{2{b^2}} \over a} = {{2(4)} \over {2\sqrt 3 }} = {4 \over {\sqrt 3 }}$$
P $$ \to $$ 4; Q $$ \to $$ 3; R $$ \to $$ 1; S $$ \to $$ 2
Hence, option (b) is correct.
Equation of hyperbola
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
It is given,
$$\angle LNM = 60^\circ {4 \over {\sqrt 3 }}$$
and Area of $$\Delta LMN = 4\sqrt 3 $$
Now, $$\Delta LMN$$ is an equilateral triangle whose sides is 2b [$$ \because $$ $$\Delta LON$$ $$ \cong $$ $$\Delta MOL$$; $$ \because $$ $$\angle NLO = \angle NMO = 60^\circ $$]
$$ \therefore $$ Area of $$\Delta LMN = {{\sqrt 3 } \over 4}{(2b)^2}$$
$$ \Rightarrow $$ $$4\sqrt 3 = \sqrt 3 {b^2} \Rightarrow b = 2$$
Also, area of $$\Delta LMN = {1 \over 2}a(2b) = ab$$
$$ \Rightarrow $$ $$4\sqrt 3 = a(2)$$
$$ \Rightarrow $$ $$a = 2\sqrt 3 $$
(P) Length of conjugate axis = 2b = 2(2) = 4 (Q)
Eccentricity (e) = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {4 \over {12}}} $$
= $${4 \over {2\sqrt 3 }} = {2 \over {\sqrt 3 }}$$
(R) Distance between the foci
= $$2ae = 2 \times 2\sqrt 3 \times {2 \over {\sqrt 3 }} = 8$$
(S) The length of latusrectum
= $${{2{b^2}} \over a} = {{2(4)} \over {2\sqrt 3 }} = {4 \over {\sqrt 3 }}$$
P $$ \to $$ 4; Q $$ \to $$ 3; R $$ \to $$ 1; S $$ \to $$ 2
Hence, option (b) is correct.
Comments (0)
