120763
Let \(S\) be the focus of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) lying on the positive \(X\) - axis and \(P\left(5, y_1\right)\) be point on the hyperbola. Then \(\mathrm{SP}=\)
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{9}{4}\)
4 \(\frac{5}{4}\)
Explanation:
C Let \(s\) be the focus of the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(a^2=16, b^2=9\)
\(a=4, b=3\)
Eccentricity,
\(\mathrm{e} =\sqrt{1+\frac{9}{16}}\)
\(\mathrm{e} =\sqrt{\frac{25}{16}}\)
\(\mathrm{e} =\frac{5}{4}\)
The point of hyperbola is \(\left(5, y_1\right)\)
\(\mathrm{SP} =\mathrm{ex}_1 \pm \mathrm{a}\)
\(=\frac{5}{4} \times 5 \pm 4=\frac{41}{4}, \frac{9}{4}\)
\(\therefore \mathrm{SP}= \frac{9}{4}\)
TS EAMCET-18.07.2022
Hyperbola
120764
The lines of the form \(x \cos \phi+y \sin \phi=\mathbf{P}\) are chords of the hyperbola \(4 x^2-y^2=4 a^2\) which subtend a right angle at the centre of the hyperbola. If these chords touch a circle with centre at \((0,0)\), then the radius of that circle is
1 \(\frac{2 a}{\sqrt{3}}\)
2 \(\frac{\mathrm{a}}{\sqrt{3}}\)
3 \(\sqrt{2 \mathrm{a}}\)
4 \(\frac{\mathrm{a}}{\sqrt{2}}\)
Explanation:
A Since, \(\mathrm{x} \cos \phi+\mathrm{y} \sin \phi=\mathrm{P}\)
Subtends a right angle at the centre \((0,0)\) of hyperbola,
\(4 \mathrm{x}^2-\mathrm{y}^2=4 \mathrm{a}^2\)
On dividing \(\left(4 a^2\right)\) both sides,
\(\frac{4 \mathrm{x}^2}{4 \mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=\frac{4 \mathrm{a}^2}{4 \mathrm{a}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=1\)
Therefore, making the hyperbola equation homogeneous with help of
\(x \cos \phi+y \sin \phi=p, \text { we get }-\)
\(\frac{x^2}{a^2}-\frac{y^2}{4 a^2}=\left(\frac{x \cos \phi+y \sin \phi}{p}\right)^2\)
\(x^2\left(\frac{1}{a^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}\right)+y^2\left(\frac{1}{4 a^2}-\frac{\sin ^2 \phi}{p^2}\right)-\frac{2 x y \cos \phi \sin \phi}{p^2}=0\)
\(\therefore\) Coefficients of \(\mathrm{x}^2+\) coefficients of \(\mathrm{y}^2=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}-\frac{1}{4 \mathrm{a}^2}-\frac{\sin ^2 \phi}{\mathrm{p}^2}=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{1}{4 \mathrm{a}^2}-\left(\frac{\cos ^2 \phi+\sin \phi}{\mathrm{p}^2}\right)=0\)
\(\left(\frac{4-1}{4 \mathrm{a}^2}\right)-\frac{1}{\mathrm{p}^2}=0 \Rightarrow \frac{3}{4 \mathrm{a}^2}=\frac{1}{\mathrm{p}^2}\)
\(\mathrm{p}^2=\frac{4 \mathrm{a}^2}{3} \Rightarrow \mathrm{p}=\frac{2}{\sqrt{3}} \mathrm{a}\)
Since, \(p\) is also length of perpendicular from \((0,0)\) to line.
\(\therefore\) Radius of circle \((\mathrm{P})=\frac{2 \mathrm{a}}{\sqrt{3}}\).
TS EAMCET-05.05.2018
Hyperbola
120765
If the latus rectum of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends an angle of \(60^{\circ}\) at the other focus, then the eccentricity of the hyperbola is
120766
If \((8,2)\) is a point on the hyperbola whose length of the transverse axis is 12 and conjugate axis is \(x=0\), then the eccentricity of that hyperbola is
1 \(\frac{2 \sqrt{2}}{7}\)
2 \(\frac{8}{5}\)
3 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
4 \(\frac{\sqrt{8}}{5}\)
Explanation:
C Let assume equation of hyperbola
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given that length of transverse axis, \(2 a=12\) or \(a=6\)
\(\therefore\) Point \((8,2)\) lies on hyperbola, then it must satisfy the equation of hyperbola
i.e., \(\frac{(8)^2}{\mathrm{a}^2}-\frac{(2)^2}{\mathrm{~b}^2}=1\)
\(\frac{64}{\mathrm{a}^2}-\frac{4}{\mathrm{~b}^2}=1\)
\(\frac{64}{(6)^2}-\frac{4}{b^2}=1\)
\(\frac{4}{\mathrm{~b}^2}=\frac{16}{9}-1=\frac{7}{9}\)
\(\mathrm{b}^2=\frac{36}{7}\)
As we know eccentricity of parabola,
\(e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{(6)^2+\frac{36}{7}}}{6}=\frac{2 \sqrt{2}}{\sqrt{7}}\)
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Hyperbola
120763
Let \(S\) be the focus of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) lying on the positive \(X\) - axis and \(P\left(5, y_1\right)\) be point on the hyperbola. Then \(\mathrm{SP}=\)
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{9}{4}\)
4 \(\frac{5}{4}\)
Explanation:
C Let \(s\) be the focus of the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(a^2=16, b^2=9\)
\(a=4, b=3\)
Eccentricity,
\(\mathrm{e} =\sqrt{1+\frac{9}{16}}\)
\(\mathrm{e} =\sqrt{\frac{25}{16}}\)
\(\mathrm{e} =\frac{5}{4}\)
The point of hyperbola is \(\left(5, y_1\right)\)
\(\mathrm{SP} =\mathrm{ex}_1 \pm \mathrm{a}\)
\(=\frac{5}{4} \times 5 \pm 4=\frac{41}{4}, \frac{9}{4}\)
\(\therefore \mathrm{SP}= \frac{9}{4}\)
TS EAMCET-18.07.2022
Hyperbola
120764
The lines of the form \(x \cos \phi+y \sin \phi=\mathbf{P}\) are chords of the hyperbola \(4 x^2-y^2=4 a^2\) which subtend a right angle at the centre of the hyperbola. If these chords touch a circle with centre at \((0,0)\), then the radius of that circle is
1 \(\frac{2 a}{\sqrt{3}}\)
2 \(\frac{\mathrm{a}}{\sqrt{3}}\)
3 \(\sqrt{2 \mathrm{a}}\)
4 \(\frac{\mathrm{a}}{\sqrt{2}}\)
Explanation:
A Since, \(\mathrm{x} \cos \phi+\mathrm{y} \sin \phi=\mathrm{P}\)
Subtends a right angle at the centre \((0,0)\) of hyperbola,
\(4 \mathrm{x}^2-\mathrm{y}^2=4 \mathrm{a}^2\)
On dividing \(\left(4 a^2\right)\) both sides,
\(\frac{4 \mathrm{x}^2}{4 \mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=\frac{4 \mathrm{a}^2}{4 \mathrm{a}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=1\)
Therefore, making the hyperbola equation homogeneous with help of
\(x \cos \phi+y \sin \phi=p, \text { we get }-\)
\(\frac{x^2}{a^2}-\frac{y^2}{4 a^2}=\left(\frac{x \cos \phi+y \sin \phi}{p}\right)^2\)
\(x^2\left(\frac{1}{a^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}\right)+y^2\left(\frac{1}{4 a^2}-\frac{\sin ^2 \phi}{p^2}\right)-\frac{2 x y \cos \phi \sin \phi}{p^2}=0\)
\(\therefore\) Coefficients of \(\mathrm{x}^2+\) coefficients of \(\mathrm{y}^2=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}-\frac{1}{4 \mathrm{a}^2}-\frac{\sin ^2 \phi}{\mathrm{p}^2}=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{1}{4 \mathrm{a}^2}-\left(\frac{\cos ^2 \phi+\sin \phi}{\mathrm{p}^2}\right)=0\)
\(\left(\frac{4-1}{4 \mathrm{a}^2}\right)-\frac{1}{\mathrm{p}^2}=0 \Rightarrow \frac{3}{4 \mathrm{a}^2}=\frac{1}{\mathrm{p}^2}\)
\(\mathrm{p}^2=\frac{4 \mathrm{a}^2}{3} \Rightarrow \mathrm{p}=\frac{2}{\sqrt{3}} \mathrm{a}\)
Since, \(p\) is also length of perpendicular from \((0,0)\) to line.
\(\therefore\) Radius of circle \((\mathrm{P})=\frac{2 \mathrm{a}}{\sqrt{3}}\).
TS EAMCET-05.05.2018
Hyperbola
120765
If the latus rectum of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends an angle of \(60^{\circ}\) at the other focus, then the eccentricity of the hyperbola is
120766
If \((8,2)\) is a point on the hyperbola whose length of the transverse axis is 12 and conjugate axis is \(x=0\), then the eccentricity of that hyperbola is
1 \(\frac{2 \sqrt{2}}{7}\)
2 \(\frac{8}{5}\)
3 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
4 \(\frac{\sqrt{8}}{5}\)
Explanation:
C Let assume equation of hyperbola
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given that length of transverse axis, \(2 a=12\) or \(a=6\)
\(\therefore\) Point \((8,2)\) lies on hyperbola, then it must satisfy the equation of hyperbola
i.e., \(\frac{(8)^2}{\mathrm{a}^2}-\frac{(2)^2}{\mathrm{~b}^2}=1\)
\(\frac{64}{\mathrm{a}^2}-\frac{4}{\mathrm{~b}^2}=1\)
\(\frac{64}{(6)^2}-\frac{4}{b^2}=1\)
\(\frac{4}{\mathrm{~b}^2}=\frac{16}{9}-1=\frac{7}{9}\)
\(\mathrm{b}^2=\frac{36}{7}\)
As we know eccentricity of parabola,
\(e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{(6)^2+\frac{36}{7}}}{6}=\frac{2 \sqrt{2}}{\sqrt{7}}\)
120763
Let \(S\) be the focus of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) lying on the positive \(X\) - axis and \(P\left(5, y_1\right)\) be point on the hyperbola. Then \(\mathrm{SP}=\)
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{9}{4}\)
4 \(\frac{5}{4}\)
Explanation:
C Let \(s\) be the focus of the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(a^2=16, b^2=9\)
\(a=4, b=3\)
Eccentricity,
\(\mathrm{e} =\sqrt{1+\frac{9}{16}}\)
\(\mathrm{e} =\sqrt{\frac{25}{16}}\)
\(\mathrm{e} =\frac{5}{4}\)
The point of hyperbola is \(\left(5, y_1\right)\)
\(\mathrm{SP} =\mathrm{ex}_1 \pm \mathrm{a}\)
\(=\frac{5}{4} \times 5 \pm 4=\frac{41}{4}, \frac{9}{4}\)
\(\therefore \mathrm{SP}= \frac{9}{4}\)
TS EAMCET-18.07.2022
Hyperbola
120764
The lines of the form \(x \cos \phi+y \sin \phi=\mathbf{P}\) are chords of the hyperbola \(4 x^2-y^2=4 a^2\) which subtend a right angle at the centre of the hyperbola. If these chords touch a circle with centre at \((0,0)\), then the radius of that circle is
1 \(\frac{2 a}{\sqrt{3}}\)
2 \(\frac{\mathrm{a}}{\sqrt{3}}\)
3 \(\sqrt{2 \mathrm{a}}\)
4 \(\frac{\mathrm{a}}{\sqrt{2}}\)
Explanation:
A Since, \(\mathrm{x} \cos \phi+\mathrm{y} \sin \phi=\mathrm{P}\)
Subtends a right angle at the centre \((0,0)\) of hyperbola,
\(4 \mathrm{x}^2-\mathrm{y}^2=4 \mathrm{a}^2\)
On dividing \(\left(4 a^2\right)\) both sides,
\(\frac{4 \mathrm{x}^2}{4 \mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=\frac{4 \mathrm{a}^2}{4 \mathrm{a}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=1\)
Therefore, making the hyperbola equation homogeneous with help of
\(x \cos \phi+y \sin \phi=p, \text { we get }-\)
\(\frac{x^2}{a^2}-\frac{y^2}{4 a^2}=\left(\frac{x \cos \phi+y \sin \phi}{p}\right)^2\)
\(x^2\left(\frac{1}{a^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}\right)+y^2\left(\frac{1}{4 a^2}-\frac{\sin ^2 \phi}{p^2}\right)-\frac{2 x y \cos \phi \sin \phi}{p^2}=0\)
\(\therefore\) Coefficients of \(\mathrm{x}^2+\) coefficients of \(\mathrm{y}^2=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}-\frac{1}{4 \mathrm{a}^2}-\frac{\sin ^2 \phi}{\mathrm{p}^2}=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{1}{4 \mathrm{a}^2}-\left(\frac{\cos ^2 \phi+\sin \phi}{\mathrm{p}^2}\right)=0\)
\(\left(\frac{4-1}{4 \mathrm{a}^2}\right)-\frac{1}{\mathrm{p}^2}=0 \Rightarrow \frac{3}{4 \mathrm{a}^2}=\frac{1}{\mathrm{p}^2}\)
\(\mathrm{p}^2=\frac{4 \mathrm{a}^2}{3} \Rightarrow \mathrm{p}=\frac{2}{\sqrt{3}} \mathrm{a}\)
Since, \(p\) is also length of perpendicular from \((0,0)\) to line.
\(\therefore\) Radius of circle \((\mathrm{P})=\frac{2 \mathrm{a}}{\sqrt{3}}\).
TS EAMCET-05.05.2018
Hyperbola
120765
If the latus rectum of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends an angle of \(60^{\circ}\) at the other focus, then the eccentricity of the hyperbola is
120766
If \((8,2)\) is a point on the hyperbola whose length of the transverse axis is 12 and conjugate axis is \(x=0\), then the eccentricity of that hyperbola is
1 \(\frac{2 \sqrt{2}}{7}\)
2 \(\frac{8}{5}\)
3 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
4 \(\frac{\sqrt{8}}{5}\)
Explanation:
C Let assume equation of hyperbola
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given that length of transverse axis, \(2 a=12\) or \(a=6\)
\(\therefore\) Point \((8,2)\) lies on hyperbola, then it must satisfy the equation of hyperbola
i.e., \(\frac{(8)^2}{\mathrm{a}^2}-\frac{(2)^2}{\mathrm{~b}^2}=1\)
\(\frac{64}{\mathrm{a}^2}-\frac{4}{\mathrm{~b}^2}=1\)
\(\frac{64}{(6)^2}-\frac{4}{b^2}=1\)
\(\frac{4}{\mathrm{~b}^2}=\frac{16}{9}-1=\frac{7}{9}\)
\(\mathrm{b}^2=\frac{36}{7}\)
As we know eccentricity of parabola,
\(e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{(6)^2+\frac{36}{7}}}{6}=\frac{2 \sqrt{2}}{\sqrt{7}}\)
120763
Let \(S\) be the focus of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) lying on the positive \(X\) - axis and \(P\left(5, y_1\right)\) be point on the hyperbola. Then \(\mathrm{SP}=\)
1 \(\frac{1}{4}\)
2 \(\frac{3}{4}\)
3 \(\frac{9}{4}\)
4 \(\frac{5}{4}\)
Explanation:
C Let \(s\) be the focus of the hyperbola,
\(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(a^2=16, b^2=9\)
\(a=4, b=3\)
Eccentricity,
\(\mathrm{e} =\sqrt{1+\frac{9}{16}}\)
\(\mathrm{e} =\sqrt{\frac{25}{16}}\)
\(\mathrm{e} =\frac{5}{4}\)
The point of hyperbola is \(\left(5, y_1\right)\)
\(\mathrm{SP} =\mathrm{ex}_1 \pm \mathrm{a}\)
\(=\frac{5}{4} \times 5 \pm 4=\frac{41}{4}, \frac{9}{4}\)
\(\therefore \mathrm{SP}= \frac{9}{4}\)
TS EAMCET-18.07.2022
Hyperbola
120764
The lines of the form \(x \cos \phi+y \sin \phi=\mathbf{P}\) are chords of the hyperbola \(4 x^2-y^2=4 a^2\) which subtend a right angle at the centre of the hyperbola. If these chords touch a circle with centre at \((0,0)\), then the radius of that circle is
1 \(\frac{2 a}{\sqrt{3}}\)
2 \(\frac{\mathrm{a}}{\sqrt{3}}\)
3 \(\sqrt{2 \mathrm{a}}\)
4 \(\frac{\mathrm{a}}{\sqrt{2}}\)
Explanation:
A Since, \(\mathrm{x} \cos \phi+\mathrm{y} \sin \phi=\mathrm{P}\)
Subtends a right angle at the centre \((0,0)\) of hyperbola,
\(4 \mathrm{x}^2-\mathrm{y}^2=4 \mathrm{a}^2\)
On dividing \(\left(4 a^2\right)\) both sides,
\(\frac{4 \mathrm{x}^2}{4 \mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=\frac{4 \mathrm{a}^2}{4 \mathrm{a}^2}\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{4 \mathrm{a}^2}=1\)
Therefore, making the hyperbola equation homogeneous with help of
\(x \cos \phi+y \sin \phi=p, \text { we get }-\)
\(\frac{x^2}{a^2}-\frac{y^2}{4 a^2}=\left(\frac{x \cos \phi+y \sin \phi}{p}\right)^2\)
\(x^2\left(\frac{1}{a^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}\right)+y^2\left(\frac{1}{4 a^2}-\frac{\sin ^2 \phi}{p^2}\right)-\frac{2 x y \cos \phi \sin \phi}{p^2}=0\)
\(\therefore\) Coefficients of \(\mathrm{x}^2+\) coefficients of \(\mathrm{y}^2=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{\cos ^2 \phi}{\mathrm{p}^2}-\frac{1}{4 \mathrm{a}^2}-\frac{\sin ^2 \phi}{\mathrm{p}^2}=0\)
\(\frac{1}{\mathrm{a}^2}-\frac{1}{4 \mathrm{a}^2}-\left(\frac{\cos ^2 \phi+\sin \phi}{\mathrm{p}^2}\right)=0\)
\(\left(\frac{4-1}{4 \mathrm{a}^2}\right)-\frac{1}{\mathrm{p}^2}=0 \Rightarrow \frac{3}{4 \mathrm{a}^2}=\frac{1}{\mathrm{p}^2}\)
\(\mathrm{p}^2=\frac{4 \mathrm{a}^2}{3} \Rightarrow \mathrm{p}=\frac{2}{\sqrt{3}} \mathrm{a}\)
Since, \(p\) is also length of perpendicular from \((0,0)\) to line.
\(\therefore\) Radius of circle \((\mathrm{P})=\frac{2 \mathrm{a}}{\sqrt{3}}\).
TS EAMCET-05.05.2018
Hyperbola
120765
If the latus rectum of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) subtends an angle of \(60^{\circ}\) at the other focus, then the eccentricity of the hyperbola is
120766
If \((8,2)\) is a point on the hyperbola whose length of the transverse axis is 12 and conjugate axis is \(x=0\), then the eccentricity of that hyperbola is
1 \(\frac{2 \sqrt{2}}{7}\)
2 \(\frac{8}{5}\)
3 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
4 \(\frac{\sqrt{8}}{5}\)
Explanation:
C Let assume equation of hyperbola
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given that length of transverse axis, \(2 a=12\) or \(a=6\)
\(\therefore\) Point \((8,2)\) lies on hyperbola, then it must satisfy the equation of hyperbola
i.e., \(\frac{(8)^2}{\mathrm{a}^2}-\frac{(2)^2}{\mathrm{~b}^2}=1\)
\(\frac{64}{\mathrm{a}^2}-\frac{4}{\mathrm{~b}^2}=1\)
\(\frac{64}{(6)^2}-\frac{4}{b^2}=1\)
\(\frac{4}{\mathrm{~b}^2}=\frac{16}{9}-1=\frac{7}{9}\)
\(\mathrm{b}^2=\frac{36}{7}\)
As we know eccentricity of parabola,
\(e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{(6)^2+\frac{36}{7}}}{6}=\frac{2 \sqrt{2}}{\sqrt{7}}\)
