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Hyperbola

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

1

\frac{\sqrt{13}}{2}

2

\frac{\sqrt{5} - 1}{2}

3

\frac{\sqrt{5} + 1}{2}

4

\frac{\sqrt{3} + 1}{2}

Explanation:

C : We have:-
Latus rectum of hyperbola subtends $90^{\circ}$ at its centre.
${Eq}^{n}$ . of hyperbola:-
original image
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
End points of latus rectum are:-
$L = (ae, \frac{b^{2}}{a}) and L^{'} = (ae, \frac{- b^{2}}{a})$
Centre $= (0, 0)$
$∵ ∠ {LCL}^{'} = 90^{\circ}$
Then, in $△ LCL$ ' by using Pythagoras theorem, $(LC)^{2} + {(L^{'} C)}^{2} = {({LL}^{'})}^{2}$
$(a e - 0)^{2} + {(\frac{b^{2}}{a} - 0)}^{2} + (a e - 0)^{2} + {(\frac{- b^{2}}{a} - 0)}^{2} = {(\frac{2 b^{2}}{a})}^{2}$
$2 (a^{2} e^{2} + \frac{b^{4}}{a^{2}}) = \frac{4 b^{4}}{a^{2}}$
$\frac{2 b^{4}}{a^{2}} = 2 a^{2} e^{2}$
$b^{4} = a^{4} e^{2}$
${[a^{2} (e^{2} - 1)]}^{2} = a^{4} e^{2} [∵ b^{2} = a^{2} (e^{2} - 1)]$
$a^{4} {(e^{2} - 1)}^{2} = a^{4} e^{2}$
${(e^{2} - 1)}^{2} = e^{2}$
$(e^{2} - 1) = \pm e$
$e^{2} \pm e - 1 = 0$
$e = \frac{\pm 1 \pm \sqrt{1 + 4}}{2} = \frac{\pm 1 \pm \sqrt{5}}{2}$
$∵$ e can't be negative,
$hence, e = \frac{\sqrt{5} + 1}{2}$

APEAPCET-20.08.2021

Hyperbola

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:,

1

\frac{8 b}{\sqrt{5}}

2

\frac{8 a}{\sqrt{5}}

3

\frac{2 a}{\sqrt{5}}

4

\frac{2 b}{\sqrt{5}}

Explanation:

(A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:, (a.) $16 x^{2} - 9 y^{2} = 16$ , (b.) $16 x^{2} - 9 y^{2} = 9$ , (c.) $16 x^{2} - 9 y^{2} = - 144$ , (d.) $16 x^{2} - 9 y^{2} = 144$ ]#
(a.) $\frac{8 b}{\sqrt{5}}$
(b.) $\frac{8 a}{\sqrt{5}}$
(c.) $\frac{2 a}{\sqrt{5}}$
(d.) $\frac{2 b}{\sqrt{5}}$
Ans: d
Exp: (D): Given that,
$e = \frac{5}{3}$
Distance between foci is 10 .
We know that,
Distance between foci $= 2 c = 10$
$c = 5$
And, $c = \sqrt{a^{2} + b^{2}} = 5$
We know, $a = \frac{c}{e} = \frac{5}{\frac{5}{3}} = 3$
From equation (i) we get-
$\sqrt{a^{2} + b^{2}} = 5$
$a^{2} + b^{2} = 25$
$(3)^{2} + b^{2} = 25$
$b^{2} = 25 - 9$
$b^{2} = 16$
$b = 4$
Hence equation of hyperbola is
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
$16 x^{2} - 9 y^{2} = 144$

AP EAPCET-24.08.2021

Hyperbola

120717 Let $x^{2} + y^{2} = 16$ be the equation of the auxiliary circle of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and let $(4 \sqrt{2, 3})$ be a point on the hyperbola. Then the eccentricity of the hyperbola is

1

5 / 4

2

5 / 3

3

4 / 3

4 2

Explanation:

A Given, $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Equation of auxiliary circle -
$x^{2} + y^{2} = 16$
We know that -
$x^{2} + y^{2} = a^{2} = 16$
And, $(4 \sqrt{2}, 3)$ point lies on hyperbola
$∴ \frac{(4 \sqrt{2})^{2}}{16} - \frac{(3)^{2}}{(b)^{2}} = 1$
$\frac{32}{16} - \frac{9}{b^{2}} = 1$
$- \frac{9}{b^{2}} = 1 - 2$
$b^{2} = 9$
$∴$ Eccentricity, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + \frac{9}{16}}$
$e = \frac{5}{4}$

AP EAMCET-06.07.2022

Hyperbola

120718 If $e_{1}, e_{2}$ are respectively the eccentricities of the curves $9 x^{2} - 16 y^{2} - 144 = 0$ and $9 x^{2} - 16 y^{2} + 144$
$= 0$ , then $\frac{e_{1}^{2} e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} =$

1

\sqrt{2}

2 1

3

\sqrt{3}

4 2

Explanation:

B Given, $C_{1} : 9 x^{2} - 16 y^{2} = 144$
$\Rightarrow\Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$C_{2} : 9 x^{2} - 16 y^{2} = - 144$
$\Rightarrow \frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
$e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{9}{16}$
$e_{1}^{2} = \frac{25}{16}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{16}{9} = \frac{25}{9}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} = \frac{25 / 16 \times 25 / 9}{\frac{25}{16} + \frac{25}{9}} = \frac{625 / 144}{625 / 144}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{1}^{2}} = 1$

AP EAMCET-20.04.2019

Hyperbola

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

1

\frac{\sqrt{13}}{2}

2

\frac{\sqrt{5} - 1}{2}

3

\frac{\sqrt{5} + 1}{2}

4

\frac{\sqrt{3} + 1}{2}

Explanation:

C : We have:-
Latus rectum of hyperbola subtends $90^{\circ}$ at its centre.
${Eq}^{n}$ . of hyperbola:-
original image
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
End points of latus rectum are:-
$L = (ae, \frac{b^{2}}{a}) and L^{'} = (ae, \frac{- b^{2}}{a})$
Centre $= (0, 0)$
$∵ ∠ {LCL}^{'} = 90^{\circ}$
Then, in $△ LCL$ ' by using Pythagoras theorem, $(LC)^{2} + {(L^{'} C)}^{2} = {({LL}^{'})}^{2}$
$(a e - 0)^{2} + {(\frac{b^{2}}{a} - 0)}^{2} + (a e - 0)^{2} + {(\frac{- b^{2}}{a} - 0)}^{2} = {(\frac{2 b^{2}}{a})}^{2}$
$2 (a^{2} e^{2} + \frac{b^{4}}{a^{2}}) = \frac{4 b^{4}}{a^{2}}$
$\frac{2 b^{4}}{a^{2}} = 2 a^{2} e^{2}$
$b^{4} = a^{4} e^{2}$
${[a^{2} (e^{2} - 1)]}^{2} = a^{4} e^{2} [∵ b^{2} = a^{2} (e^{2} - 1)]$
$a^{4} {(e^{2} - 1)}^{2} = a^{4} e^{2}$
${(e^{2} - 1)}^{2} = e^{2}$
$(e^{2} - 1) = \pm e$
$e^{2} \pm e - 1 = 0$
$e = \frac{\pm 1 \pm \sqrt{1 + 4}}{2} = \frac{\pm 1 \pm \sqrt{5}}{2}$
$∵$ e can't be negative,
$hence, e = \frac{\sqrt{5} + 1}{2}$

APEAPCET-20.08.2021

Hyperbola

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:,

1

\frac{8 b}{\sqrt{5}}

2

\frac{8 a}{\sqrt{5}}

3

\frac{2 a}{\sqrt{5}}

4

\frac{2 b}{\sqrt{5}}

Explanation:

(A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:, (a.) $16 x^{2} - 9 y^{2} = 16$ , (b.) $16 x^{2} - 9 y^{2} = 9$ , (c.) $16 x^{2} - 9 y^{2} = - 144$ , (d.) $16 x^{2} - 9 y^{2} = 144$ ]#
(a.) $\frac{8 b}{\sqrt{5}}$
(b.) $\frac{8 a}{\sqrt{5}}$
(c.) $\frac{2 a}{\sqrt{5}}$
(d.) $\frac{2 b}{\sqrt{5}}$
Ans: d
Exp: (D): Given that,
$e = \frac{5}{3}$
Distance between foci is 10 .
We know that,
Distance between foci $= 2 c = 10$
$c = 5$
And, $c = \sqrt{a^{2} + b^{2}} = 5$
We know, $a = \frac{c}{e} = \frac{5}{\frac{5}{3}} = 3$
From equation (i) we get-
$\sqrt{a^{2} + b^{2}} = 5$
$a^{2} + b^{2} = 25$
$(3)^{2} + b^{2} = 25$
$b^{2} = 25 - 9$
$b^{2} = 16$
$b = 4$
Hence equation of hyperbola is
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
$16 x^{2} - 9 y^{2} = 144$

AP EAPCET-24.08.2021

Hyperbola

120717 Let $x^{2} + y^{2} = 16$ be the equation of the auxiliary circle of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and let $(4 \sqrt{2, 3})$ be a point on the hyperbola. Then the eccentricity of the hyperbola is

1

5 / 4

2

5 / 3

3

4 / 3

4 2

Explanation:

A Given, $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Equation of auxiliary circle -
$x^{2} + y^{2} = 16$
We know that -
$x^{2} + y^{2} = a^{2} = 16$
And, $(4 \sqrt{2}, 3)$ point lies on hyperbola
$∴ \frac{(4 \sqrt{2})^{2}}{16} - \frac{(3)^{2}}{(b)^{2}} = 1$
$\frac{32}{16} - \frac{9}{b^{2}} = 1$
$- \frac{9}{b^{2}} = 1 - 2$
$b^{2} = 9$
$∴$ Eccentricity, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + \frac{9}{16}}$
$e = \frac{5}{4}$

AP EAMCET-06.07.2022

Hyperbola

120718 If $e_{1}, e_{2}$ are respectively the eccentricities of the curves $9 x^{2} - 16 y^{2} - 144 = 0$ and $9 x^{2} - 16 y^{2} + 144$
$= 0$ , then $\frac{e_{1}^{2} e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} =$

1

\sqrt{2}

2 1

3

\sqrt{3}

4 2

Explanation:

B Given, $C_{1} : 9 x^{2} - 16 y^{2} = 144$
$\Rightarrow\Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$C_{2} : 9 x^{2} - 16 y^{2} = - 144$
$\Rightarrow \frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
$e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{9}{16}$
$e_{1}^{2} = \frac{25}{16}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{16}{9} = \frac{25}{9}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} = \frac{25 / 16 \times 25 / 9}{\frac{25}{16} + \frac{25}{9}} = \frac{625 / 144}{625 / 144}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{1}^{2}} = 1$

AP EAMCET-20.04.2019

Hyperbola

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

1

\frac{\sqrt{13}}{2}

2

\frac{\sqrt{5} - 1}{2}

3

\frac{\sqrt{5} + 1}{2}

4

\frac{\sqrt{3} + 1}{2}

Explanation:

C : We have:-
Latus rectum of hyperbola subtends $90^{\circ}$ at its centre.
${Eq}^{n}$ . of hyperbola:-
original image
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
End points of latus rectum are:-
$L = (ae, \frac{b^{2}}{a}) and L^{'} = (ae, \frac{- b^{2}}{a})$
Centre $= (0, 0)$
$∵ ∠ {LCL}^{'} = 90^{\circ}$
Then, in $△ LCL$ ' by using Pythagoras theorem, $(LC)^{2} + {(L^{'} C)}^{2} = {({LL}^{'})}^{2}$
$(a e - 0)^{2} + {(\frac{b^{2}}{a} - 0)}^{2} + (a e - 0)^{2} + {(\frac{- b^{2}}{a} - 0)}^{2} = {(\frac{2 b^{2}}{a})}^{2}$
$2 (a^{2} e^{2} + \frac{b^{4}}{a^{2}}) = \frac{4 b^{4}}{a^{2}}$
$\frac{2 b^{4}}{a^{2}} = 2 a^{2} e^{2}$
$b^{4} = a^{4} e^{2}$
${[a^{2} (e^{2} - 1)]}^{2} = a^{4} e^{2} [∵ b^{2} = a^{2} (e^{2} - 1)]$
$a^{4} {(e^{2} - 1)}^{2} = a^{4} e^{2}$
${(e^{2} - 1)}^{2} = e^{2}$
$(e^{2} - 1) = \pm e$
$e^{2} \pm e - 1 = 0$
$e = \frac{\pm 1 \pm \sqrt{1 + 4}}{2} = \frac{\pm 1 \pm \sqrt{5}}{2}$
$∵$ e can't be negative,
$hence, e = \frac{\sqrt{5} + 1}{2}$

APEAPCET-20.08.2021

Hyperbola

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:,

1

\frac{8 b}{\sqrt{5}}

2

\frac{8 a}{\sqrt{5}}

3

\frac{2 a}{\sqrt{5}}

4

\frac{2 b}{\sqrt{5}}

Explanation:

(A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:, (a.) $16 x^{2} - 9 y^{2} = 16$ , (b.) $16 x^{2} - 9 y^{2} = 9$ , (c.) $16 x^{2} - 9 y^{2} = - 144$ , (d.) $16 x^{2} - 9 y^{2} = 144$ ]#
(a.) $\frac{8 b}{\sqrt{5}}$
(b.) $\frac{8 a}{\sqrt{5}}$
(c.) $\frac{2 a}{\sqrt{5}}$
(d.) $\frac{2 b}{\sqrt{5}}$
Ans: d
Exp: (D): Given that,
$e = \frac{5}{3}$
Distance between foci is 10 .
We know that,
Distance between foci $= 2 c = 10$
$c = 5$
And, $c = \sqrt{a^{2} + b^{2}} = 5$
We know, $a = \frac{c}{e} = \frac{5}{\frac{5}{3}} = 3$
From equation (i) we get-
$\sqrt{a^{2} + b^{2}} = 5$
$a^{2} + b^{2} = 25$
$(3)^{2} + b^{2} = 25$
$b^{2} = 25 - 9$
$b^{2} = 16$
$b = 4$
Hence equation of hyperbola is
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
$16 x^{2} - 9 y^{2} = 144$

AP EAPCET-24.08.2021

Hyperbola

120717 Let $x^{2} + y^{2} = 16$ be the equation of the auxiliary circle of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and let $(4 \sqrt{2, 3})$ be a point on the hyperbola. Then the eccentricity of the hyperbola is

1

5 / 4

2

5 / 3

3

4 / 3

4 2

Explanation:

A Given, $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Equation of auxiliary circle -
$x^{2} + y^{2} = 16$
We know that -
$x^{2} + y^{2} = a^{2} = 16$
And, $(4 \sqrt{2}, 3)$ point lies on hyperbola
$∴ \frac{(4 \sqrt{2})^{2}}{16} - \frac{(3)^{2}}{(b)^{2}} = 1$
$\frac{32}{16} - \frac{9}{b^{2}} = 1$
$- \frac{9}{b^{2}} = 1 - 2$
$b^{2} = 9$
$∴$ Eccentricity, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + \frac{9}{16}}$
$e = \frac{5}{4}$

AP EAMCET-06.07.2022

Hyperbola

120718 If $e_{1}, e_{2}$ are respectively the eccentricities of the curves $9 x^{2} - 16 y^{2} - 144 = 0$ and $9 x^{2} - 16 y^{2} + 144$
$= 0$ , then $\frac{e_{1}^{2} e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} =$

1

\sqrt{2}

2 1

3

\sqrt{3}

4 2

Explanation:

B Given, $C_{1} : 9 x^{2} - 16 y^{2} = 144$
$\Rightarrow\Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$C_{2} : 9 x^{2} - 16 y^{2} = - 144$
$\Rightarrow \frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
$e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{9}{16}$
$e_{1}^{2} = \frac{25}{16}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{16}{9} = \frac{25}{9}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} = \frac{25 / 16 \times 25 / 9}{\frac{25}{16} + \frac{25}{9}} = \frac{625 / 144}{625 / 144}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{1}^{2}} = 1$

AP EAMCET-20.04.2019

Hyperbola

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

1

\frac{\sqrt{13}}{2}

2

\frac{\sqrt{5} - 1}{2}

3

\frac{\sqrt{5} + 1}{2}

4

\frac{\sqrt{3} + 1}{2}

Explanation:

C : We have:-
Latus rectum of hyperbola subtends $90^{\circ}$ at its centre.
${Eq}^{n}$ . of hyperbola:-
original image
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
End points of latus rectum are:-
$L = (ae, \frac{b^{2}}{a}) and L^{'} = (ae, \frac{- b^{2}}{a})$
Centre $= (0, 0)$
$∵ ∠ {LCL}^{'} = 90^{\circ}$
Then, in $△ LCL$ ' by using Pythagoras theorem, $(LC)^{2} + {(L^{'} C)}^{2} = {({LL}^{'})}^{2}$
$(a e - 0)^{2} + {(\frac{b^{2}}{a} - 0)}^{2} + (a e - 0)^{2} + {(\frac{- b^{2}}{a} - 0)}^{2} = {(\frac{2 b^{2}}{a})}^{2}$
$2 (a^{2} e^{2} + \frac{b^{4}}{a^{2}}) = \frac{4 b^{4}}{a^{2}}$
$\frac{2 b^{4}}{a^{2}} = 2 a^{2} e^{2}$
$b^{4} = a^{4} e^{2}$
${[a^{2} (e^{2} - 1)]}^{2} = a^{4} e^{2} [∵ b^{2} = a^{2} (e^{2} - 1)]$
$a^{4} {(e^{2} - 1)}^{2} = a^{4} e^{2}$
${(e^{2} - 1)}^{2} = e^{2}$
$(e^{2} - 1) = \pm e$
$e^{2} \pm e - 1 = 0$
$e = \frac{\pm 1 \pm \sqrt{1 + 4}}{2} = \frac{\pm 1 \pm \sqrt{5}}{2}$
$∵$ e can't be negative,
$hence, e = \frac{\sqrt{5} + 1}{2}$

APEAPCET-20.08.2021

Hyperbola

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:,

1

\frac{8 b}{\sqrt{5}}

2

\frac{8 a}{\sqrt{5}}

3

\frac{2 a}{\sqrt{5}}

4

\frac{2 b}{\sqrt{5}}

Explanation:

(A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , $∵$ Length of transverse axis $= 2 \times$ length of conjugate axis., $2 a = 2 \times 2 b \Rightarrow a = 2 b$ , $∵ b^{2} = a^{2} (e^{2} - 1)$ , $\frac{a^{2}}{4} = a^{2} (e^{2} - 1)$ , $e^{2} = 1 + \frac{1}{4} = \frac{5}{4}$ , $e = \frac{\sqrt{5}}{2}$ , $∴$ Distance between its directrix $= \frac{2 a}{e} = \frac{2 (2 b)}{\frac{\sqrt{5}}{2}} = \frac{8 b}{\sqrt{5}}$ , 998. The equation of hyperbola whose eccentricity is $\frac{5}{3}$ and distance between the foci is 10 units is:, (a.) $16 x^{2} - 9 y^{2} = 16$ , (b.) $16 x^{2} - 9 y^{2} = 9$ , (c.) $16 x^{2} - 9 y^{2} = - 144$ , (d.) $16 x^{2} - 9 y^{2} = 144$ ]#
(a.) $\frac{8 b}{\sqrt{5}}$
(b.) $\frac{8 a}{\sqrt{5}}$
(c.) $\frac{2 a}{\sqrt{5}}$
(d.) $\frac{2 b}{\sqrt{5}}$
Ans: d
Exp: (D): Given that,
$e = \frac{5}{3}$
Distance between foci is 10 .
We know that,
Distance between foci $= 2 c = 10$
$c = 5$
And, $c = \sqrt{a^{2} + b^{2}} = 5$
We know, $a = \frac{c}{e} = \frac{5}{\frac{5}{3}} = 3$
From equation (i) we get-
$\sqrt{a^{2} + b^{2}} = 5$
$a^{2} + b^{2} = 25$
$(3)^{2} + b^{2} = 25$
$b^{2} = 25 - 9$
$b^{2} = 16$
$b = 4$
Hence equation of hyperbola is
$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
$16 x^{2} - 9 y^{2} = 144$

AP EAPCET-24.08.2021

Hyperbola

120717 Let $x^{2} + y^{2} = 16$ be the equation of the auxiliary circle of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and let $(4 \sqrt{2, 3})$ be a point on the hyperbola. Then the eccentricity of the hyperbola is

1

5 / 4

2

5 / 3

3

4 / 3

4 2

Explanation:

A Given, $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Equation of auxiliary circle -
$x^{2} + y^{2} = 16$
We know that -
$x^{2} + y^{2} = a^{2} = 16$
And, $(4 \sqrt{2}, 3)$ point lies on hyperbola
$∴ \frac{(4 \sqrt{2})^{2}}{16} - \frac{(3)^{2}}{(b)^{2}} = 1$
$\frac{32}{16} - \frac{9}{b^{2}} = 1$
$- \frac{9}{b^{2}} = 1 - 2$
$b^{2} = 9$
$∴$ Eccentricity, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + \frac{9}{16}}$
$e = \frac{5}{4}$

AP EAMCET-06.07.2022

Hyperbola

120718 If $e_{1}, e_{2}$ are respectively the eccentricities of the curves $9 x^{2} - 16 y^{2} - 144 = 0$ and $9 x^{2} - 16 y^{2} + 144$
$= 0$ , then $\frac{e_{1}^{2} e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} =$

1

\sqrt{2}

2 1

3

\sqrt{3}

4 2

Explanation:

B Given, $C_{1} : 9 x^{2} - 16 y^{2} = 144$
$\Rightarrow\Rightarrow \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$C_{2} : 9 x^{2} - 16 y^{2} = - 144$
$\Rightarrow \frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$
$e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{9}{16}$
$e_{1}^{2} = \frac{25}{16}$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{16}{9} = \frac{25}{9}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{2}^{2}} = \frac{25 / 16 \times 25 / 9}{\frac{25}{16} + \frac{25}{9}} = \frac{625 / 144}{625 / 144}$
$\frac{e_{1}^{2} \cdot e_{2}^{2}}{e_{1}^{2} + e_{1}^{2}} = 1$

AP EAMCET-20.04.2019