120832
The product of distances from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) to its two asymptotes is
1 \(\frac{25}{144}\)
2 \(\frac{144}{25}\)
3 \(\frac{12}{5}\)
4 None of these
Explanation:
B Given equation is hyperbola,
\(\frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=1\)
On comparing standerd hyperbola equation,
\(a^2=16 \text { and } b^2=9\)
Where,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(d=\frac{a^2 b^2}{a^2+b^2}=\frac{16 \times 9}{16+9}\)
\(d=\frac{144}{25}\)
EAMCET-1993
Hyperbola
120833
If the equation of one asymptote of the hyperbola \(14 x^2+38 x y+20 y^2+x-7 y-91=0\) is \(7 x+5 y-3=0\), then the other asymptote is
1 \(2 x-4 y+1=0\)
2 \(2 x+4 y+1=0\)
3 \(2 \mathrm{x}-4 \mathrm{y}-1=0\)
4 \(2 x+4 y-1=0\)
Explanation:
B Given,
\(14 x^2+38 x y+20 y^2+x-7 y-91=0\)
We know that, euqation of the pair of line -
\(\Delta=a b c+2 f g h-a f^2-b^2-c h^2=0\)
Here, \(\mathrm{a}=14, \mathrm{~b}=20, \mathrm{f}=19, \mathrm{~g}=\frac{1}{2}, \mathrm{~h}=\frac{-7}{2}\)
\(14(20) \mathrm{c}+2(19)\left(\frac{1}{2}\right)\left(\frac{-7}{2}\right)-14\left(\frac{7}{2}\right)^2-20\left(\frac{1}{2}\right)^2-\mathrm{c}(19)^2=0\)
\(81 \mathrm{c}=-243\)
\(\mathrm{c}=-3\)
\(\therefore 14 x^2+38 x y+20 y^2+x-7 y-3=(7 x+5 y-3)(2 x+\) \(4 y+1)\).
So, other asymptote is :
\(2 \mathrm{x}+4 \mathrm{y}+1=0\)
AP EAMCET-24.04.2018
Hyperbola
120834
The asymptotes of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), with any tangent to the hyperbola form a triangle whose area is \(a^2 \tan (\alpha)\). Then its eccentricity equals
1 \(\sec (\alpha)\)
2 \(\operatorname{cosec}(\alpha)\)
3 \(\sec ^2(\alpha)\)
4 \(\operatorname{cosec}^2(\alpha)\)
Explanation:
A : Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to \((a b)\).
\(\mathrm{a}^2 \tan (\alpha)=\mathrm{ab}\)
\(\tan \alpha=\frac{\mathrm{b}}{\mathrm{a}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \quad[\mathrm{e}>>1]\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1+(\tan \alpha)^2}\)
\(\mathrm{e}=\sqrt{\sec ^2 \alpha}\)
\(\mathrm{e}=\sec \alpha\)
AP EAMCET-19.08.2021
Hyperbola
120835
The equation of the hyperbola whose asymptotes are the lines \(3 x+4 y-2=0\), \(2 \mathrm{x}+\mathrm{y}+\mathbf{1}=\mathbf{0}\) and which passes through the point \((1,1)\) is
1 \(6 x^2+11 x y+4 y^2-30 x+2 y+7=0\)
2 \(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
3 \(6 x^2+11 x y+4 y^2-x+2 y+22=0\)
4 \(6 x^2+11 x y+4 y^2-3 x-7 y-11=0\)
Explanation:
B Given,
\(l_1: 3 \mathrm{x}+4 \mathrm{y}-2=0\)
\(l_2: 2 \mathrm{x}+\mathrm{y}+1+0\)
Equation of hyperbola,
\((3 \mathrm{x}+4 \mathrm{y}-2)(2 \mathrm{x}+\mathrm{y}+1)+\mathrm{k}=0\)
On passing at point \((1,1)\) we get -
\((3+4-2)(2+1+1)+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
Since, it is passes through the point \((1,1)\)
Then,
\((3 \times 1+4 \times 1-2)(2 \times 1+1+1)+\mathrm{k}=0\)
\(5 \times 4+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
\(\mathrm{k}=-20\)
Hence, equation of hyperbola is -
\(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hyperbola
120832
The product of distances from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) to its two asymptotes is
1 \(\frac{25}{144}\)
2 \(\frac{144}{25}\)
3 \(\frac{12}{5}\)
4 None of these
Explanation:
B Given equation is hyperbola,
\(\frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=1\)
On comparing standerd hyperbola equation,
\(a^2=16 \text { and } b^2=9\)
Where,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(d=\frac{a^2 b^2}{a^2+b^2}=\frac{16 \times 9}{16+9}\)
\(d=\frac{144}{25}\)
EAMCET-1993
Hyperbola
120833
If the equation of one asymptote of the hyperbola \(14 x^2+38 x y+20 y^2+x-7 y-91=0\) is \(7 x+5 y-3=0\), then the other asymptote is
1 \(2 x-4 y+1=0\)
2 \(2 x+4 y+1=0\)
3 \(2 \mathrm{x}-4 \mathrm{y}-1=0\)
4 \(2 x+4 y-1=0\)
Explanation:
B Given,
\(14 x^2+38 x y+20 y^2+x-7 y-91=0\)
We know that, euqation of the pair of line -
\(\Delta=a b c+2 f g h-a f^2-b^2-c h^2=0\)
Here, \(\mathrm{a}=14, \mathrm{~b}=20, \mathrm{f}=19, \mathrm{~g}=\frac{1}{2}, \mathrm{~h}=\frac{-7}{2}\)
\(14(20) \mathrm{c}+2(19)\left(\frac{1}{2}\right)\left(\frac{-7}{2}\right)-14\left(\frac{7}{2}\right)^2-20\left(\frac{1}{2}\right)^2-\mathrm{c}(19)^2=0\)
\(81 \mathrm{c}=-243\)
\(\mathrm{c}=-3\)
\(\therefore 14 x^2+38 x y+20 y^2+x-7 y-3=(7 x+5 y-3)(2 x+\) \(4 y+1)\).
So, other asymptote is :
\(2 \mathrm{x}+4 \mathrm{y}+1=0\)
AP EAMCET-24.04.2018
Hyperbola
120834
The asymptotes of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), with any tangent to the hyperbola form a triangle whose area is \(a^2 \tan (\alpha)\). Then its eccentricity equals
1 \(\sec (\alpha)\)
2 \(\operatorname{cosec}(\alpha)\)
3 \(\sec ^2(\alpha)\)
4 \(\operatorname{cosec}^2(\alpha)\)
Explanation:
A : Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to \((a b)\).
\(\mathrm{a}^2 \tan (\alpha)=\mathrm{ab}\)
\(\tan \alpha=\frac{\mathrm{b}}{\mathrm{a}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \quad[\mathrm{e}>>1]\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1+(\tan \alpha)^2}\)
\(\mathrm{e}=\sqrt{\sec ^2 \alpha}\)
\(\mathrm{e}=\sec \alpha\)
AP EAMCET-19.08.2021
Hyperbola
120835
The equation of the hyperbola whose asymptotes are the lines \(3 x+4 y-2=0\), \(2 \mathrm{x}+\mathrm{y}+\mathbf{1}=\mathbf{0}\) and which passes through the point \((1,1)\) is
1 \(6 x^2+11 x y+4 y^2-30 x+2 y+7=0\)
2 \(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
3 \(6 x^2+11 x y+4 y^2-x+2 y+22=0\)
4 \(6 x^2+11 x y+4 y^2-3 x-7 y-11=0\)
Explanation:
B Given,
\(l_1: 3 \mathrm{x}+4 \mathrm{y}-2=0\)
\(l_2: 2 \mathrm{x}+\mathrm{y}+1+0\)
Equation of hyperbola,
\((3 \mathrm{x}+4 \mathrm{y}-2)(2 \mathrm{x}+\mathrm{y}+1)+\mathrm{k}=0\)
On passing at point \((1,1)\) we get -
\((3+4-2)(2+1+1)+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
Since, it is passes through the point \((1,1)\)
Then,
\((3 \times 1+4 \times 1-2)(2 \times 1+1+1)+\mathrm{k}=0\)
\(5 \times 4+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
\(\mathrm{k}=-20\)
Hence, equation of hyperbola is -
\(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
120832
The product of distances from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) to its two asymptotes is
1 \(\frac{25}{144}\)
2 \(\frac{144}{25}\)
3 \(\frac{12}{5}\)
4 None of these
Explanation:
B Given equation is hyperbola,
\(\frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=1\)
On comparing standerd hyperbola equation,
\(a^2=16 \text { and } b^2=9\)
Where,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(d=\frac{a^2 b^2}{a^2+b^2}=\frac{16 \times 9}{16+9}\)
\(d=\frac{144}{25}\)
EAMCET-1993
Hyperbola
120833
If the equation of one asymptote of the hyperbola \(14 x^2+38 x y+20 y^2+x-7 y-91=0\) is \(7 x+5 y-3=0\), then the other asymptote is
1 \(2 x-4 y+1=0\)
2 \(2 x+4 y+1=0\)
3 \(2 \mathrm{x}-4 \mathrm{y}-1=0\)
4 \(2 x+4 y-1=0\)
Explanation:
B Given,
\(14 x^2+38 x y+20 y^2+x-7 y-91=0\)
We know that, euqation of the pair of line -
\(\Delta=a b c+2 f g h-a f^2-b^2-c h^2=0\)
Here, \(\mathrm{a}=14, \mathrm{~b}=20, \mathrm{f}=19, \mathrm{~g}=\frac{1}{2}, \mathrm{~h}=\frac{-7}{2}\)
\(14(20) \mathrm{c}+2(19)\left(\frac{1}{2}\right)\left(\frac{-7}{2}\right)-14\left(\frac{7}{2}\right)^2-20\left(\frac{1}{2}\right)^2-\mathrm{c}(19)^2=0\)
\(81 \mathrm{c}=-243\)
\(\mathrm{c}=-3\)
\(\therefore 14 x^2+38 x y+20 y^2+x-7 y-3=(7 x+5 y-3)(2 x+\) \(4 y+1)\).
So, other asymptote is :
\(2 \mathrm{x}+4 \mathrm{y}+1=0\)
AP EAMCET-24.04.2018
Hyperbola
120834
The asymptotes of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), with any tangent to the hyperbola form a triangle whose area is \(a^2 \tan (\alpha)\). Then its eccentricity equals
1 \(\sec (\alpha)\)
2 \(\operatorname{cosec}(\alpha)\)
3 \(\sec ^2(\alpha)\)
4 \(\operatorname{cosec}^2(\alpha)\)
Explanation:
A : Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to \((a b)\).
\(\mathrm{a}^2 \tan (\alpha)=\mathrm{ab}\)
\(\tan \alpha=\frac{\mathrm{b}}{\mathrm{a}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \quad[\mathrm{e}>>1]\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1+(\tan \alpha)^2}\)
\(\mathrm{e}=\sqrt{\sec ^2 \alpha}\)
\(\mathrm{e}=\sec \alpha\)
AP EAMCET-19.08.2021
Hyperbola
120835
The equation of the hyperbola whose asymptotes are the lines \(3 x+4 y-2=0\), \(2 \mathrm{x}+\mathrm{y}+\mathbf{1}=\mathbf{0}\) and which passes through the point \((1,1)\) is
1 \(6 x^2+11 x y+4 y^2-30 x+2 y+7=0\)
2 \(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
3 \(6 x^2+11 x y+4 y^2-x+2 y+22=0\)
4 \(6 x^2+11 x y+4 y^2-3 x-7 y-11=0\)
Explanation:
B Given,
\(l_1: 3 \mathrm{x}+4 \mathrm{y}-2=0\)
\(l_2: 2 \mathrm{x}+\mathrm{y}+1+0\)
Equation of hyperbola,
\((3 \mathrm{x}+4 \mathrm{y}-2)(2 \mathrm{x}+\mathrm{y}+1)+\mathrm{k}=0\)
On passing at point \((1,1)\) we get -
\((3+4-2)(2+1+1)+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
Since, it is passes through the point \((1,1)\)
Then,
\((3 \times 1+4 \times 1-2)(2 \times 1+1+1)+\mathrm{k}=0\)
\(5 \times 4+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
\(\mathrm{k}=-20\)
Hence, equation of hyperbola is -
\(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
120832
The product of distances from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) to its two asymptotes is
1 \(\frac{25}{144}\)
2 \(\frac{144}{25}\)
3 \(\frac{12}{5}\)
4 None of these
Explanation:
B Given equation is hyperbola,
\(\frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=1\)
On comparing standerd hyperbola equation,
\(a^2=16 \text { and } b^2=9\)
Where,
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(d=\frac{a^2 b^2}{a^2+b^2}=\frac{16 \times 9}{16+9}\)
\(d=\frac{144}{25}\)
EAMCET-1993
Hyperbola
120833
If the equation of one asymptote of the hyperbola \(14 x^2+38 x y+20 y^2+x-7 y-91=0\) is \(7 x+5 y-3=0\), then the other asymptote is
1 \(2 x-4 y+1=0\)
2 \(2 x+4 y+1=0\)
3 \(2 \mathrm{x}-4 \mathrm{y}-1=0\)
4 \(2 x+4 y-1=0\)
Explanation:
B Given,
\(14 x^2+38 x y+20 y^2+x-7 y-91=0\)
We know that, euqation of the pair of line -
\(\Delta=a b c+2 f g h-a f^2-b^2-c h^2=0\)
Here, \(\mathrm{a}=14, \mathrm{~b}=20, \mathrm{f}=19, \mathrm{~g}=\frac{1}{2}, \mathrm{~h}=\frac{-7}{2}\)
\(14(20) \mathrm{c}+2(19)\left(\frac{1}{2}\right)\left(\frac{-7}{2}\right)-14\left(\frac{7}{2}\right)^2-20\left(\frac{1}{2}\right)^2-\mathrm{c}(19)^2=0\)
\(81 \mathrm{c}=-243\)
\(\mathrm{c}=-3\)
\(\therefore 14 x^2+38 x y+20 y^2+x-7 y-3=(7 x+5 y-3)(2 x+\) \(4 y+1)\).
So, other asymptote is :
\(2 \mathrm{x}+4 \mathrm{y}+1=0\)
AP EAMCET-24.04.2018
Hyperbola
120834
The asymptotes of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), with any tangent to the hyperbola form a triangle whose area is \(a^2 \tan (\alpha)\). Then its eccentricity equals
1 \(\sec (\alpha)\)
2 \(\operatorname{cosec}(\alpha)\)
3 \(\sec ^2(\alpha)\)
4 \(\operatorname{cosec}^2(\alpha)\)
Explanation:
A : Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to \((a b)\).
\(\mathrm{a}^2 \tan (\alpha)=\mathrm{ab}\)
\(\tan \alpha=\frac{\mathrm{b}}{\mathrm{a}}\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \quad[\mathrm{e}>>1]\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1+(\tan \alpha)^2}\)
\(\mathrm{e}=\sqrt{\sec ^2 \alpha}\)
\(\mathrm{e}=\sec \alpha\)
AP EAMCET-19.08.2021
Hyperbola
120835
The equation of the hyperbola whose asymptotes are the lines \(3 x+4 y-2=0\), \(2 \mathrm{x}+\mathrm{y}+\mathbf{1}=\mathbf{0}\) and which passes through the point \((1,1)\) is
1 \(6 x^2+11 x y+4 y^2-30 x+2 y+7=0\)
2 \(6 x^2+11 x y+4 y^2-x+2 y-22=0\)
3 \(6 x^2+11 x y+4 y^2-x+2 y+22=0\)
4 \(6 x^2+11 x y+4 y^2-3 x-7 y-11=0\)
Explanation:
B Given,
\(l_1: 3 \mathrm{x}+4 \mathrm{y}-2=0\)
\(l_2: 2 \mathrm{x}+\mathrm{y}+1+0\)
Equation of hyperbola,
\((3 \mathrm{x}+4 \mathrm{y}-2)(2 \mathrm{x}+\mathrm{y}+1)+\mathrm{k}=0\)
On passing at point \((1,1)\) we get -
\((3+4-2)(2+1+1)+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
Since, it is passes through the point \((1,1)\)
Then,
\((3 \times 1+4 \times 1-2)(2 \times 1+1+1)+\mathrm{k}=0\)
\(5 \times 4+\mathrm{k}=0\)
\(20+\mathrm{k}=0\)
\(\mathrm{k}=-20\)
Hence, equation of hyperbola is -
\(6 x^2+11 x y+4 y^2-x+2 y-22=0\)