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Hyperbola

120831 The product of the perpendicular distance from any point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ to its asymptotes is.

\frac{a^{2} b^{2}}{a^{2} - b^{2}}

\frac{a^{2} b^{2}}{a^{2} + b^{2}}

\frac{a^{2} + b^{2}}{a^{2} b^{2}}

\frac{a^{2} - b^{2}}{a^{2} b^{2}}

Explanation:

B Given equation is hyperbola,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
The product of perpendicular distance from any point on this hyperbola to its asymptotes is.
$d = \frac{a^{2} b^{2}}{a^{2} + b^{2}}$

AP EAMCET-2010

Hyperbola

120840 A rectangular hyperbola passing through $(3, 2)$ has its asymptotes parallel to the coordinate axes. If $(1, 1)$ is the point of intersection of the two perpendicular tangents of that hyperbola, then its equation is

x y = x + \frac{1}{y}

x (y + 1 + \frac{1}{x}) = 1

x (1 - y) = y - 1

x y = x + y + 1

Explanation:

D It is given that a rectangular hyperbola whose asymptotes are parallel to coordinate axis and point of intersection of perpendicular tangents (mean centre of hyperbola) is $(1, 1)$ , the equation of hyperbola we can take as
$(x - 1) (y - 1) = k^{2}$
$∵$ Hyperbola (i) passes through point $(3, 2)$ , so $k = 2$ Therefore equation of the required hyperbola is
$(x - 1) (y - 1) = 2 \Rightarrow x y = x + y + 1$

TS EAMCET-11.09.2020

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Hyperbola

120831 The product of the perpendicular distance from any point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ to its asymptotes is.

\frac{a^{2} b^{2}}{a^{2} - b^{2}}

\frac{a^{2} b^{2}}{a^{2} + b^{2}}

\frac{a^{2} + b^{2}}{a^{2} b^{2}}

\frac{a^{2} - b^{2}}{a^{2} b^{2}}

Explanation:

AP EAMCET-2010

Hyperbola

x y = x + \frac{1}{y}

x (y + 1 + \frac{1}{x}) = 1

x (1 - y) = y - 1

x y = x + y + 1

Explanation:

TS EAMCET-11.09.2020

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