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.
1 \(\frac{a^2 b^2}{a^2-b^2}\)
2 \(\frac{a^2 b^2}{a^2+b^2}\)
3 \(\frac{a^2+b^2}{a^2 b^2}\)
4 \(\frac{a^2-b^2}{a^2 b^2}\)
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.
\(\mathrm{d}=\frac{\mathrm{a}^2 \mathrm{~b}^2}{\mathrm{a}^2+\mathrm{b}^2}\)
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
1 \(x y=x+\frac{1}{y}\)
2 \(x\left(y+1+\frac{1}{x}\right)=1\)
3 \(x(1-y)=y-1\)
4 \(x y=x+y+1\)
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
\((\mathrm{x}-1)(\mathrm{y}-1)=\mathrm{k}^2\)
\(\because\) Hyperbola (i) passes through point \((3,2)\), so \(\mathrm{k}=2\) Therefore equation of the required hyperbola is
\((x-1)(y-1)=2 \Rightarrow x y=x+y+1\)
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.
1 \(\frac{a^2 b^2}{a^2-b^2}\)
2 \(\frac{a^2 b^2}{a^2+b^2}\)
3 \(\frac{a^2+b^2}{a^2 b^2}\)
4 \(\frac{a^2-b^2}{a^2 b^2}\)
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.
\(\mathrm{d}=\frac{\mathrm{a}^2 \mathrm{~b}^2}{\mathrm{a}^2+\mathrm{b}^2}\)
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
1 \(x y=x+\frac{1}{y}\)
2 \(x\left(y+1+\frac{1}{x}\right)=1\)
3 \(x(1-y)=y-1\)
4 \(x y=x+y+1\)
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
\((\mathrm{x}-1)(\mathrm{y}-1)=\mathrm{k}^2\)
\(\because\) Hyperbola (i) passes through point \((3,2)\), so \(\mathrm{k}=2\) Therefore equation of the required hyperbola is
\((x-1)(y-1)=2 \Rightarrow x y=x+y+1\)
