Let the foci and length of the latus rectum of an ellipse , a b be (5, 0) and , respectively. Then, the square of the eccentricity of the hyperbola equals [2024]

Question

Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

Smart Achievers · Accepted Answer

(51)

Here ae = 5 and $\frac{2 b^{2}}{a} = \sqrt{50}$

$\frac{b^{2}}{a} = \frac{5 \sqrt{2}}{2}$

$a^{2} e^{2} = a^{2} - b^{2} \Rightarrow e^{2} = 1 - \frac{b^{2}}{a^{2}}$

$25 = a^{2} - \frac{5 \sqrt{2}}{2} a \Rightarrow 2 a^{2} - 5 \sqrt{2} a - 50 = 0$

$\Rightarrow [a - 5 \sqrt{2}] [2 a + 5 \sqrt{2}] = 0 \Rightarrow a = 5 \sqrt{2} or a = \frac{- 5 \sqrt{2}}{2}$

$\Rightarrow e = \frac{5}{a} = \frac{1}{\sqrt{2}}$

Also, $b^{2} = 5 \sqrt{2} \times \frac{5 \sqrt{2}}{2} = 25 \Rightarrow b = 5$

Now, $e = \sqrt{1 + \frac{a^{2} b^{2}}{b^{2}}} = \sqrt{1 + a^{2}} \Rightarrow e^{2} = 1 + a^{2} = 51$ .

Solution

Q.
Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]

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Solution

Q. Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1, a > b be (±5, 0) and 50, respectively. Then, the square of the eccentricity of the hyperbola x2b2–y2a2b2=1 equals [2024]

Ans. (51) Here ae = 5 and 2b2a=50 b2a=522 a2e2=a2–b2 ⇒ e2=1–b2a2 25=a2–522a ⇒ 2a2–52a–50=0 ⇒ [a–52][2a+52]=0 ⇒ a=52 or a=–522 ⇒ e=5a=12 Also, b2=52×522=25 ⇒ b=5 Now, e=1+a2b2b2=1+a2 ⇒ e2=1+a2=51.

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Q.
Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b be ( $\pm$ 5, 0) and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2} b^{2}} = 1$ equals [2024]