120719
The locus of a variable point whose chord of contact w.r.t the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends a right angle at the origin is
D Given, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the pole of chord (PQ)
\(\therefore\) equation of polar \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1\)
The centre of hyperbola is \((0,0)\)
Let it be the point \(c\), making equation (i) - (ii) we get-
\(\therefore \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=\left(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}\right)\)
Point \(P\) subtends at \(90^{\circ}\) with \((0,0)\)
\(\therefore\) Lines \(\mathrm{CP}\) and \(\mathrm{CQ}\) is given by (ii) are perpendicular -
\(\therefore \quad\left(\frac{1}{\mathrm{a}^2}-\frac{\mathrm{x}_1^2}{\mathrm{a}^4}\right)+\left(\frac{-1}{\mathrm{~b}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}\right)=0\)
\(\frac{\mathrm{x}_1^2}{\mathrm{a}^4}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^4}+\frac{\mathrm{y}^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
AP EAMCET-04.07.2021
Hyperbola
120720
The equation of the transverse axis of hyperbola \((x-3)^2+(y+1)^2=(4 x+3 y)^2\) is
1 \(3 x+4 y=13\)
2 \(3 x-4 y=13\)
3 \(4 x-3 y=13\)
4 \(3 x-4 y=9\)
Explanation:
B Given, \((x-3)^2+(y+1)^2=(4 x+3 y)^2\)
\(\Rightarrow \quad(\mathrm{x}-3)^2+(\mathrm{y}+1)^2=25\left(\frac{4 \mathrm{x}+3 \mathrm{y}}{5}\right)^2\)
\(\mathrm{PS}=5 \mathrm{MP}\).
Directrix is
\(4 x+3 y=0 \& \text { focus }(3,-1)\)
Slope, of traverse axis is -
\(\frac{-1}{-4 / 3}=\frac{3}{4}\)
Equation of traverse axis -
\((y+1)=\frac{3}{4}(x-3)\)
\(3 x-4 y=13\)
AP EAMCET-18.09.2020
Hyperbola
120721
If the eccentricity of a hyperbola is \(\sqrt{3}\); then the eccentricity of its conjugate hyperbola is:
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(\sqrt{\frac{3}{2}}\)
4 \(2 \sqrt{3}\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{3}\)
Let, \(\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Conjugate hyperbola; \(\mathrm{H}^{\prime}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{3} .\)
\(3=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=2\)
\(\frac{\mathrm{a}^2}{\mathrm{~b}^2}=\frac{1}{2}\)
\(\mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
\(\mathrm{e}^{\prime}=\sqrt{\frac{3}{2}}\)Now, \(\quad \mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
AP EAMCET-2006
Hyperbola
120722
The eccentricity of the hyperbola \(9 x^2-16 y^2+\) \(72 x-32 y-16=0\) is
120723
A hyperbola passing through a focus of the ellipse \(\frac{\mathbf{x}^2}{169}+\frac{y^2}{25}=1\). Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then the equation of the hyperbola is
1 \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
2 \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
3 \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Explanation:
C Given,
\(\mathrm{E}: \frac{\mathrm{x}^2}{169}+\frac{\mathrm{y}^2}{25}=1\)
Here, \(\quad \mathrm{a}=13\) and \(\mathrm{b}=5\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{169}} \quad \mathrm{e}=\frac{12}{13}\)
Eccentricity of hyperbola is e,
\(\therefore \quad\) e \(\cdot \mathrm{e}_1=1\)
\(\mathrm{e}_1=\frac{1}{\mathrm{e}}=\frac{13}{12}\)
\(\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}_1^2}}=\frac{13}{12}\)
\(\frac{\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)}{\mathrm{a}_1^2}=\frac{169}{144}\)
\(\mathrm{a}_1^2=144 \&\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)=169\)
\(\mathrm{b}_1^2=169-\mathrm{a}_1^2=169-144\)
\(\mathrm{b}_1^2=25\)
Here, equation of hyperbola is
\(\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1\)
\(\frac{x^2}{144}-\frac{y^2}{25}=1\)
120719
The locus of a variable point whose chord of contact w.r.t the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends a right angle at the origin is
D Given, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the pole of chord (PQ)
\(\therefore\) equation of polar \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1\)
The centre of hyperbola is \((0,0)\)
Let it be the point \(c\), making equation (i) - (ii) we get-
\(\therefore \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=\left(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}\right)\)
Point \(P\) subtends at \(90^{\circ}\) with \((0,0)\)
\(\therefore\) Lines \(\mathrm{CP}\) and \(\mathrm{CQ}\) is given by (ii) are perpendicular -
\(\therefore \quad\left(\frac{1}{\mathrm{a}^2}-\frac{\mathrm{x}_1^2}{\mathrm{a}^4}\right)+\left(\frac{-1}{\mathrm{~b}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}\right)=0\)
\(\frac{\mathrm{x}_1^2}{\mathrm{a}^4}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^4}+\frac{\mathrm{y}^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
AP EAMCET-04.07.2021
Hyperbola
120720
The equation of the transverse axis of hyperbola \((x-3)^2+(y+1)^2=(4 x+3 y)^2\) is
1 \(3 x+4 y=13\)
2 \(3 x-4 y=13\)
3 \(4 x-3 y=13\)
4 \(3 x-4 y=9\)
Explanation:
B Given, \((x-3)^2+(y+1)^2=(4 x+3 y)^2\)
\(\Rightarrow \quad(\mathrm{x}-3)^2+(\mathrm{y}+1)^2=25\left(\frac{4 \mathrm{x}+3 \mathrm{y}}{5}\right)^2\)
\(\mathrm{PS}=5 \mathrm{MP}\).
Directrix is
\(4 x+3 y=0 \& \text { focus }(3,-1)\)
Slope, of traverse axis is -
\(\frac{-1}{-4 / 3}=\frac{3}{4}\)
Equation of traverse axis -
\((y+1)=\frac{3}{4}(x-3)\)
\(3 x-4 y=13\)
AP EAMCET-18.09.2020
Hyperbola
120721
If the eccentricity of a hyperbola is \(\sqrt{3}\); then the eccentricity of its conjugate hyperbola is:
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(\sqrt{\frac{3}{2}}\)
4 \(2 \sqrt{3}\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{3}\)
Let, \(\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Conjugate hyperbola; \(\mathrm{H}^{\prime}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{3} .\)
\(3=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=2\)
\(\frac{\mathrm{a}^2}{\mathrm{~b}^2}=\frac{1}{2}\)
\(\mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
\(\mathrm{e}^{\prime}=\sqrt{\frac{3}{2}}\)Now, \(\quad \mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
AP EAMCET-2006
Hyperbola
120722
The eccentricity of the hyperbola \(9 x^2-16 y^2+\) \(72 x-32 y-16=0\) is
120723
A hyperbola passing through a focus of the ellipse \(\frac{\mathbf{x}^2}{169}+\frac{y^2}{25}=1\). Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then the equation of the hyperbola is
1 \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
2 \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
3 \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Explanation:
C Given,
\(\mathrm{E}: \frac{\mathrm{x}^2}{169}+\frac{\mathrm{y}^2}{25}=1\)
Here, \(\quad \mathrm{a}=13\) and \(\mathrm{b}=5\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{169}} \quad \mathrm{e}=\frac{12}{13}\)
Eccentricity of hyperbola is e,
\(\therefore \quad\) e \(\cdot \mathrm{e}_1=1\)
\(\mathrm{e}_1=\frac{1}{\mathrm{e}}=\frac{13}{12}\)
\(\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}_1^2}}=\frac{13}{12}\)
\(\frac{\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)}{\mathrm{a}_1^2}=\frac{169}{144}\)
\(\mathrm{a}_1^2=144 \&\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)=169\)
\(\mathrm{b}_1^2=169-\mathrm{a}_1^2=169-144\)
\(\mathrm{b}_1^2=25\)
Here, equation of hyperbola is
\(\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1\)
\(\frac{x^2}{144}-\frac{y^2}{25}=1\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120719
The locus of a variable point whose chord of contact w.r.t the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends a right angle at the origin is
D Given, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the pole of chord (PQ)
\(\therefore\) equation of polar \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1\)
The centre of hyperbola is \((0,0)\)
Let it be the point \(c\), making equation (i) - (ii) we get-
\(\therefore \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=\left(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}\right)\)
Point \(P\) subtends at \(90^{\circ}\) with \((0,0)\)
\(\therefore\) Lines \(\mathrm{CP}\) and \(\mathrm{CQ}\) is given by (ii) are perpendicular -
\(\therefore \quad\left(\frac{1}{\mathrm{a}^2}-\frac{\mathrm{x}_1^2}{\mathrm{a}^4}\right)+\left(\frac{-1}{\mathrm{~b}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}\right)=0\)
\(\frac{\mathrm{x}_1^2}{\mathrm{a}^4}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^4}+\frac{\mathrm{y}^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
AP EAMCET-04.07.2021
Hyperbola
120720
The equation of the transverse axis of hyperbola \((x-3)^2+(y+1)^2=(4 x+3 y)^2\) is
1 \(3 x+4 y=13\)
2 \(3 x-4 y=13\)
3 \(4 x-3 y=13\)
4 \(3 x-4 y=9\)
Explanation:
B Given, \((x-3)^2+(y+1)^2=(4 x+3 y)^2\)
\(\Rightarrow \quad(\mathrm{x}-3)^2+(\mathrm{y}+1)^2=25\left(\frac{4 \mathrm{x}+3 \mathrm{y}}{5}\right)^2\)
\(\mathrm{PS}=5 \mathrm{MP}\).
Directrix is
\(4 x+3 y=0 \& \text { focus }(3,-1)\)
Slope, of traverse axis is -
\(\frac{-1}{-4 / 3}=\frac{3}{4}\)
Equation of traverse axis -
\((y+1)=\frac{3}{4}(x-3)\)
\(3 x-4 y=13\)
AP EAMCET-18.09.2020
Hyperbola
120721
If the eccentricity of a hyperbola is \(\sqrt{3}\); then the eccentricity of its conjugate hyperbola is:
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(\sqrt{\frac{3}{2}}\)
4 \(2 \sqrt{3}\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{3}\)
Let, \(\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Conjugate hyperbola; \(\mathrm{H}^{\prime}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{3} .\)
\(3=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=2\)
\(\frac{\mathrm{a}^2}{\mathrm{~b}^2}=\frac{1}{2}\)
\(\mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
\(\mathrm{e}^{\prime}=\sqrt{\frac{3}{2}}\)Now, \(\quad \mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
AP EAMCET-2006
Hyperbola
120722
The eccentricity of the hyperbola \(9 x^2-16 y^2+\) \(72 x-32 y-16=0\) is
120723
A hyperbola passing through a focus of the ellipse \(\frac{\mathbf{x}^2}{169}+\frac{y^2}{25}=1\). Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then the equation of the hyperbola is
1 \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
2 \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
3 \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Explanation:
C Given,
\(\mathrm{E}: \frac{\mathrm{x}^2}{169}+\frac{\mathrm{y}^2}{25}=1\)
Here, \(\quad \mathrm{a}=13\) and \(\mathrm{b}=5\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{169}} \quad \mathrm{e}=\frac{12}{13}\)
Eccentricity of hyperbola is e,
\(\therefore \quad\) e \(\cdot \mathrm{e}_1=1\)
\(\mathrm{e}_1=\frac{1}{\mathrm{e}}=\frac{13}{12}\)
\(\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}_1^2}}=\frac{13}{12}\)
\(\frac{\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)}{\mathrm{a}_1^2}=\frac{169}{144}\)
\(\mathrm{a}_1^2=144 \&\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)=169\)
\(\mathrm{b}_1^2=169-\mathrm{a}_1^2=169-144\)
\(\mathrm{b}_1^2=25\)
Here, equation of hyperbola is
\(\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1\)
\(\frac{x^2}{144}-\frac{y^2}{25}=1\)
120719
The locus of a variable point whose chord of contact w.r.t the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends a right angle at the origin is
D Given, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the pole of chord (PQ)
\(\therefore\) equation of polar \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1\)
The centre of hyperbola is \((0,0)\)
Let it be the point \(c\), making equation (i) - (ii) we get-
\(\therefore \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=\left(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}\right)\)
Point \(P\) subtends at \(90^{\circ}\) with \((0,0)\)
\(\therefore\) Lines \(\mathrm{CP}\) and \(\mathrm{CQ}\) is given by (ii) are perpendicular -
\(\therefore \quad\left(\frac{1}{\mathrm{a}^2}-\frac{\mathrm{x}_1^2}{\mathrm{a}^4}\right)+\left(\frac{-1}{\mathrm{~b}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}\right)=0\)
\(\frac{\mathrm{x}_1^2}{\mathrm{a}^4}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^4}+\frac{\mathrm{y}^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
AP EAMCET-04.07.2021
Hyperbola
120720
The equation of the transverse axis of hyperbola \((x-3)^2+(y+1)^2=(4 x+3 y)^2\) is
1 \(3 x+4 y=13\)
2 \(3 x-4 y=13\)
3 \(4 x-3 y=13\)
4 \(3 x-4 y=9\)
Explanation:
B Given, \((x-3)^2+(y+1)^2=(4 x+3 y)^2\)
\(\Rightarrow \quad(\mathrm{x}-3)^2+(\mathrm{y}+1)^2=25\left(\frac{4 \mathrm{x}+3 \mathrm{y}}{5}\right)^2\)
\(\mathrm{PS}=5 \mathrm{MP}\).
Directrix is
\(4 x+3 y=0 \& \text { focus }(3,-1)\)
Slope, of traverse axis is -
\(\frac{-1}{-4 / 3}=\frac{3}{4}\)
Equation of traverse axis -
\((y+1)=\frac{3}{4}(x-3)\)
\(3 x-4 y=13\)
AP EAMCET-18.09.2020
Hyperbola
120721
If the eccentricity of a hyperbola is \(\sqrt{3}\); then the eccentricity of its conjugate hyperbola is:
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(\sqrt{\frac{3}{2}}\)
4 \(2 \sqrt{3}\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{3}\)
Let, \(\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Conjugate hyperbola; \(\mathrm{H}^{\prime}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{3} .\)
\(3=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=2\)
\(\frac{\mathrm{a}^2}{\mathrm{~b}^2}=\frac{1}{2}\)
\(\mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
\(\mathrm{e}^{\prime}=\sqrt{\frac{3}{2}}\)Now, \(\quad \mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
AP EAMCET-2006
Hyperbola
120722
The eccentricity of the hyperbola \(9 x^2-16 y^2+\) \(72 x-32 y-16=0\) is
120723
A hyperbola passing through a focus of the ellipse \(\frac{\mathbf{x}^2}{169}+\frac{y^2}{25}=1\). Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then the equation of the hyperbola is
1 \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
2 \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
3 \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Explanation:
C Given,
\(\mathrm{E}: \frac{\mathrm{x}^2}{169}+\frac{\mathrm{y}^2}{25}=1\)
Here, \(\quad \mathrm{a}=13\) and \(\mathrm{b}=5\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{169}} \quad \mathrm{e}=\frac{12}{13}\)
Eccentricity of hyperbola is e,
\(\therefore \quad\) e \(\cdot \mathrm{e}_1=1\)
\(\mathrm{e}_1=\frac{1}{\mathrm{e}}=\frac{13}{12}\)
\(\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}_1^2}}=\frac{13}{12}\)
\(\frac{\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)}{\mathrm{a}_1^2}=\frac{169}{144}\)
\(\mathrm{a}_1^2=144 \&\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)=169\)
\(\mathrm{b}_1^2=169-\mathrm{a}_1^2=169-144\)
\(\mathrm{b}_1^2=25\)
Here, equation of hyperbola is
\(\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1\)
\(\frac{x^2}{144}-\frac{y^2}{25}=1\)
120719
The locus of a variable point whose chord of contact w.r.t the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends a right angle at the origin is
D Given, \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) be the pole of chord (PQ)
\(\therefore\) equation of polar \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) is -
\(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1\)
The centre of hyperbola is \((0,0)\)
Let it be the point \(c\), making equation (i) - (ii) we get-
\(\therefore \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=\left(\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}\right)\)
Point \(P\) subtends at \(90^{\circ}\) with \((0,0)\)
\(\therefore\) Lines \(\mathrm{CP}\) and \(\mathrm{CQ}\) is given by (ii) are perpendicular -
\(\therefore \quad\left(\frac{1}{\mathrm{a}^2}-\frac{\mathrm{x}_1^2}{\mathrm{a}^4}\right)+\left(\frac{-1}{\mathrm{~b}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}\right)=0\)
\(\frac{\mathrm{x}_1^2}{\mathrm{a}^4}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^4}+\frac{\mathrm{y}^2}{\mathrm{~b}^4}=\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
AP EAMCET-04.07.2021
Hyperbola
120720
The equation of the transverse axis of hyperbola \((x-3)^2+(y+1)^2=(4 x+3 y)^2\) is
1 \(3 x+4 y=13\)
2 \(3 x-4 y=13\)
3 \(4 x-3 y=13\)
4 \(3 x-4 y=9\)
Explanation:
B Given, \((x-3)^2+(y+1)^2=(4 x+3 y)^2\)
\(\Rightarrow \quad(\mathrm{x}-3)^2+(\mathrm{y}+1)^2=25\left(\frac{4 \mathrm{x}+3 \mathrm{y}}{5}\right)^2\)
\(\mathrm{PS}=5 \mathrm{MP}\).
Directrix is
\(4 x+3 y=0 \& \text { focus }(3,-1)\)
Slope, of traverse axis is -
\(\frac{-1}{-4 / 3}=\frac{3}{4}\)
Equation of traverse axis -
\((y+1)=\frac{3}{4}(x-3)\)
\(3 x-4 y=13\)
AP EAMCET-18.09.2020
Hyperbola
120721
If the eccentricity of a hyperbola is \(\sqrt{3}\); then the eccentricity of its conjugate hyperbola is:
1 \(\sqrt{2}\)
2 \(\sqrt{3}\)
3 \(\sqrt{\frac{3}{2}}\)
4 \(2 \sqrt{3}\)
Explanation:
C Given, \(\mathrm{e}=\sqrt{3}\)
Let, \(\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Conjugate hyperbola; \(\mathrm{H}^{\prime}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1\)
\(\mathrm{e}=\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{3} .\)
\(3=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{\mathrm{b}^2}{\mathrm{a}^2}=2\)
\(\frac{\mathrm{a}^2}{\mathrm{~b}^2}=\frac{1}{2}\)
\(\mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
\(\mathrm{e}^{\prime}=\sqrt{\frac{3}{2}}\)Now, \(\quad \mathrm{e}^{\prime}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{1+\left(\frac{1}{2}\right)}\)
AP EAMCET-2006
Hyperbola
120722
The eccentricity of the hyperbola \(9 x^2-16 y^2+\) \(72 x-32 y-16=0\) is
120723
A hyperbola passing through a focus of the ellipse \(\frac{\mathbf{x}^2}{169}+\frac{y^2}{25}=1\). Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then the equation of the hyperbola is
1 \(\frac{x^2}{144}-\frac{y^2}{9}=1\)
2 \(\frac{x^2}{169}-\frac{y^2}{25}=1\)
3 \(\frac{x^2}{144}-\frac{y^2}{25}=1\)
4 \(\frac{x^2}{25}-\frac{y^2}{9}=1\)
Explanation:
C Given,
\(\mathrm{E}: \frac{\mathrm{x}^2}{169}+\frac{\mathrm{y}^2}{25}=1\)
Here, \(\quad \mathrm{a}=13\) and \(\mathrm{b}=5\)
\(\therefore \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{169}} \quad \mathrm{e}=\frac{12}{13}\)
Eccentricity of hyperbola is e,
\(\therefore \quad\) e \(\cdot \mathrm{e}_1=1\)
\(\mathrm{e}_1=\frac{1}{\mathrm{e}}=\frac{13}{12}\)
\(\sqrt{1+\frac{\mathrm{b}^2}{\mathrm{a}_1^2}}=\frac{13}{12}\)
\(\frac{\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)}{\mathrm{a}_1^2}=\frac{169}{144}\)
\(\mathrm{a}_1^2=144 \&\left(\mathrm{a}_1^2+\mathrm{b}_1^2\right)=169\)
\(\mathrm{b}_1^2=169-\mathrm{a}_1^2=169-144\)
\(\mathrm{b}_1^2=25\)
Here, equation of hyperbola is
\(\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1\)
\(\frac{x^2}{144}-\frac{y^2}{25}=1\)