The equation of a hyperbola is (x^2)/(a^2)-(y^2)/(b^2)=1. If P√2,√5 is a point from which perpendicular tangents can be drawn to the hyperbola and…

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The equation of a hyperbola is (x^2)/(a^2)-(y^2)/(b^2)=1. If P√2,√5 is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is 10, then its eccentricity is

MARKS Content Team · Accepted Answer

Correct answer: (B) 5/4. ∵P lies on director circle x^2+y^2=a^2-b^2 ∴ a^2-b^2=7 …(i) Also, 2ae=10⇒ae=5 …(ii) Now, e^2=(a^2+b^2)/(a^2)⇒a^2+b^2=25 …(iii) From (i) & (iii), we get, a^2=16,b^2=9 e=√2516=5/4

The equation of a hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If $P (\sqrt{2}, \sqrt{5})$ is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is $10$ , then its eccentricity is

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The equation of a hyperbola is x2a2-y2b2=1. If P2,5 is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is 10, then its eccentricity is

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The equation of a hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If $P (\sqrt{2}, \sqrt{5})$ is a point from which perpendicular tangents can be drawn to the hyperbola and distance between both the foci of the hyperbola is $10$ , then its eccentricity is