120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), \(\because\) Length of transverse axis \(=2 \times\) length of conjugate axis., \(2 \mathrm{a} =2 \times 2 \mathrm{~b} \Rightarrow \mathrm{a}=2 \mathrm{~b}\), \(\because \quad \mathrm{b}^2 =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\frac{\mathrm{a}^2}{4} =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\mathrm{e}^2 =1+\frac{1}{4}=\frac{5}{4}\), \(\mathrm{e} =\frac{\sqrt{5}}{2}\), \(\therefore\) Distance between its directrix \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(2 \mathrm{~b})}{\frac{\sqrt{5}}{2}}=\frac{8 \mathrm{~b}}{\sqrt{5}}\), 998. The equation of hyperbola whose eccentricity is \(\frac{5}{3}\) and distance between the foci is 10 units is:,

120717 Let \(x^2+y^2=16\) be the equation of the auxiliary circle of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and let \((4 \sqrt{2,3})\) be a point on the hyperbola. Then the eccentricity of the hyperbola is

120718 If \(e_1, e_2\) are respectively the eccentricities of the curves \(9 x^2-16 y^2-144=0\) and \(9 x^2-16 y^2+144\)
\(=0\), then \(\frac{\mathbf{e}_1^2 \mathrm{e}_2^2}{\mathrm{e}_1^2+\mathrm{e}_2^2}=\)

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), \(\because\) Length of transverse axis \(=2 \times\) length of conjugate axis., \(2 \mathrm{a} =2 \times 2 \mathrm{~b} \Rightarrow \mathrm{a}=2 \mathrm{~b}\), \(\because \quad \mathrm{b}^2 =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\frac{\mathrm{a}^2}{4} =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\mathrm{e}^2 =1+\frac{1}{4}=\frac{5}{4}\), \(\mathrm{e} =\frac{\sqrt{5}}{2}\), \(\therefore\) Distance between its directrix \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(2 \mathrm{~b})}{\frac{\sqrt{5}}{2}}=\frac{8 \mathrm{~b}}{\sqrt{5}}\), 998. The equation of hyperbola whose eccentricity is \(\frac{5}{3}\) and distance between the foci is 10 units is:,

120717 Let \(x^2+y^2=16\) be the equation of the auxiliary circle of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and let \((4 \sqrt{2,3})\) be a point on the hyperbola. Then the eccentricity of the hyperbola is

120718 If \(e_1, e_2\) are respectively the eccentricities of the curves \(9 x^2-16 y^2-144=0\) and \(9 x^2-16 y^2+144\)
\(=0\), then \(\frac{\mathbf{e}_1^2 \mathrm{e}_2^2}{\mathrm{e}_1^2+\mathrm{e}_2^2}=\)

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), \(\because\) Length of transverse axis \(=2 \times\) length of conjugate axis., \(2 \mathrm{a} =2 \times 2 \mathrm{~b} \Rightarrow \mathrm{a}=2 \mathrm{~b}\), \(\because \quad \mathrm{b}^2 =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\frac{\mathrm{a}^2}{4} =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\mathrm{e}^2 =1+\frac{1}{4}=\frac{5}{4}\), \(\mathrm{e} =\frac{\sqrt{5}}{2}\), \(\therefore\) Distance between its directrix \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(2 \mathrm{~b})}{\frac{\sqrt{5}}{2}}=\frac{8 \mathrm{~b}}{\sqrt{5}}\), 998. The equation of hyperbola whose eccentricity is \(\frac{5}{3}\) and distance between the foci is 10 units is:,

120717 Let \(x^2+y^2=16\) be the equation of the auxiliary circle of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and let \((4 \sqrt{2,3})\) be a point on the hyperbola. Then the eccentricity of the hyperbola is

120718 If \(e_1, e_2\) are respectively the eccentricities of the curves \(9 x^2-16 y^2-144=0\) and \(9 x^2-16 y^2+144\)
\(=0\), then \(\frac{\mathbf{e}_1^2 \mathrm{e}_2^2}{\mathrm{e}_1^2+\mathrm{e}_2^2}=\)

120715 If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is:

120716 In a hyperbola if the length of transverse axis is twice that of the conjugate axis. then the distance between its directrices is units.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET-24.08.2021,Shift-II], Exp: (A): Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), \(\because\) Length of transverse axis \(=2 \times\) length of conjugate axis., \(2 \mathrm{a} =2 \times 2 \mathrm{~b} \Rightarrow \mathrm{a}=2 \mathrm{~b}\), \(\because \quad \mathrm{b}^2 =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\frac{\mathrm{a}^2}{4} =\mathrm{a}^2\left(\mathrm{e}^2-1\right)\), \(\mathrm{e}^2 =1+\frac{1}{4}=\frac{5}{4}\), \(\mathrm{e} =\frac{\sqrt{5}}{2}\), \(\therefore\) Distance between its directrix \(=\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(2 \mathrm{~b})}{\frac{\sqrt{5}}{2}}=\frac{8 \mathrm{~b}}{\sqrt{5}}\), 998. The equation of hyperbola whose eccentricity is \(\frac{5}{3}\) and distance between the foci is 10 units is:,

120717 Let \(x^2+y^2=16\) be the equation of the auxiliary circle of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and let \((4 \sqrt{2,3})\) be a point on the hyperbola. Then the eccentricity of the hyperbola is

120718 If \(e_1, e_2\) are respectively the eccentricities of the curves \(9 x^2-16 y^2-144=0\) and \(9 x^2-16 y^2+144\)
\(=0\), then \(\frac{\mathbf{e}_1^2 \mathrm{e}_2^2}{\mathrm{e}_1^2+\mathrm{e}_2^2}=\)