For the hyperbola x 2 100 - y 2 25 = 1 , prove that S A · S ' A = 25 , where S and S ' are the foci and A is the vertex.

Given equation of the hyperbola: x 2 100 - y 2 25 = 1 . On comparing with the standard equation of the hyperbola x 2 a 2 - y 2 b 2 = 1 , we get a 2 = 100 , b 2 = 25 ∴ a = 10 , b = 5 and we know that, b 2 = a 2 e 2 - 1 ∴ 25 = 100 e 2 - 1 ⇒ 1 4 = e 2 - 1 ⇒ 5 4 = e 2 ⇒ e = 5 2 (Since eccentricity is non-negative) So, the foci of the hyperbola is S = a e , 0 = 10 × 5 2 , 0 = 5 5 , 0 Therefore, S = 5 5 , 0 , S ' = - 5 5 , 0 and vertex A = a , 0 = 10 , 0 Now, S A = 5 5 - 10 2 and S ' A = - 5 5 - 10 2 ∵ distance = x 2 - x 1 2 + y 2 - y 1 2 ∴ S A · S ' A = ( 5 5 - 10 ) 2 · ( 5 5 + 10 ) 2 ⇒ S A · S ' A = ( 5 5 - 10 ) · ( 5 5 + 10 ) ⇒ S A · S ' A = ( 125 - 100 ) ⇒ S A · S ' A = 25 Hence, proved.

Find the equation of the hyperbola in the standard form if eccentricity is 3 2 and distance between foci is 12

Let the required equation of hyperbola be x 2 a 2 - y 2 b 2 = 1 Given, eccentricity ( e ) = 3 2 Distance between foci = 2 a e Given, distance between foci = 12 ⇒ 2 a e = 12 ⇒ 2 a 3 2 = 12 3 a = 12 ⇒ a = 12 3 = 4 a 2 = 16 Now, b 2 = a 2 e 2 - 1 ⇒ b 2 = 16 3 2 2 - 1 = 16 9 4 - 1 = 16 5 4 ⇒ b 2 = 20 ∴ The required equation of hyperbola is x 2 16 - y 2 20 = 1

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Mathematics and Statistics (Arts & Science) Part 1>Chapter 7 - Conic Sections>MISCELLANEOUS EXERCISE - 7>Q 15

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Find the eccentricity of an ellipse if the distance between its directrices is three times the distance between its foci.

Important Questions on Conic Sections

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Mathematics and Statistics (Arts & Science) Part 1>Chapter 7 - Conic Sections>MISCELLANEOUS EXERCISE - 7>Q 16

For the hyperbola $\frac{x^{2}}{100} - \frac{y^{2}}{25} = 1$ , prove that $S A \cdot S^{'} A = 25,$ where $S$ and $S^{'}$ are the foci and $A$ is the vertex.