SATHEE: Hyperbola 1 Question 2

Hyperbola 1 Question 2

3.

If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 2 \sqrt{3})$ is $5 x = 4 \sqrt{5}$ and its eccentricity is $e$ , then

(2019 Main, 10 April I)

(a) $4 e^{4} - 12 e^{2} - 27 = 0$

(b) $4 e^{4} - 24 e^{2} + 27 = 0$

(d) $4 e^{4} - 24 e^{2} + 35 = 0$

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Answer:

Correct Answer: 3. (d)

Solution:

Let the equation of hyperbola is

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Since, equation of given directrix is $5 x = 4 \sqrt{5}$

$\begin{aligned} so & 5 \frac{a}{e} & = 4 \sqrt{5} [∵ equation of directrix is x = \frac{a}{e}] \\ \Rightarrow & \frac{a}{e} = \frac{4}{\sqrt{5}} \end{aligned}$

and hyperbola (i) passes through point $(4, - 2 \sqrt{3})$

$so, \frac{16}{a^{2}} - \frac{12}{b^{2}} = 1$

The eccentricity $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$

$\begin{array}{lrl} \Rightarrow & e^{2} = 1 + \frac{b^{2}}{a^{2}} \\ \Rightarrow & a^{2} e^{2} - a^{2} = b^{2} \end{array}$

From Eqs. (ii) and (iv), we get

$\frac{16}{5} e^{4} - \frac{16}{5} e^{2} = b^{2}$

From Eqs. (ii) and (iii), we get

$\begin{aligned} \frac{16}{\frac{16}{5} e^{2}} - \frac{12}{b^{2}} = 1 \Rightarrow \frac{5}{e^{2}} - \frac{12}{b^{2}} = 1 \\ \Rightarrow & \frac{12}{b^{2}} = \frac{5}{e^{2}} - 1 \Rightarrow \frac{12}{b^{2}} = \frac{5 - e^{2}}{e^{2}} \\ \Rightarrow & b^{2} = \frac{12 e^{2}}{5 - e^{2}} \end{aligned}$

From Eqs. (v) and (vi), we get

$\begin{aligned} 16 e^{4} - 16 e^{2} = 5 \frac{12 e^{2}}{5 - e^{2}} \Rightarrow 16 (e^{2} - 1) (5 - e^{2}) = 60 \\ \Rightarrow & 4 (5 e^{2} - e^{4} - 5 + e^{2}) = 15 \\ \Rightarrow & 4 e^{4} - 24 e^{2} + 35 = 0 \end{aligned}$

Hyperbola 1 Question 2

3.

Answer:

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Hyperbola 1 Question 2

3.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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