NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120789
The locus of the mid-points of the chords of the circle \(x^2+y^2=16\) which are the tangents to the hyperbola \(9 x^2-16 y^2=144\) is
1 \(3 x^2-4 y^2=16\left(x^2+y^2\right)\)
2 \(4 x^2-3 y^2=9\left(x^2+y^2\right)\)
3 \(16 x^2-9 y^2=\left(x^2+y^2\right)^2\)
4 \(16 x^2-9 y^2=4\left(x^2+y^2\right)\)
Explanation:
C Given,
\(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { be a point in the locus }\)
\(\mathrm{x}^2+\mathrm{y}^2=16 \text { having } \mathrm{P} \text { is its midpoint }\)
\(\mathrm{xx}_1+\mathrm{yy}_1-16=\mathrm{x}_1^2+\mathrm{y}_1^2-16\)
\(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\)
If (1) is a tangent to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(16\left(\mathrm{x}_1\right)^2-9\left(\mathrm{y}_1\right)^2=\left(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\right)^2\)\(\therefore\) The locus of P is \(16 \mathrm{x}^2-9 \mathrm{y}^2=\left(\mathrm{x}^2+\mathrm{y}^2\right)^2\)
AP EAMCET-21.04.2019
Hyperbola
120790
The values of \(\mathrm{m}\) for which the line \(\mathrm{y}=\mathrm{mx}+2\) becomes a tangent to the hyperbola \(4 \mathrm{x}^2-9 \mathrm{y}^2=\) 36 is
1 \(\pm \frac{2}{3}\)
2 \(\pm \frac{2 \sqrt{2}}{3}\)
3 \(\pm \frac{8}{9}\)
4 \(\pm \frac{4 \sqrt{2}}{3}\)
Explanation:
B Given, \(4 x^2-9 y^2=36\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
We know that for a tangent to the hyperbola,
\(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\((2)^2=9 \mathrm{~m}^2-4\)
\(\therefore \quad(2)^2=9 \mathrm{~m}^2-4 \quad\{\) from, \(\mathrm{y}=\mathrm{mx}+2\}\)
\(\Rightarrow \mathrm{m}= \pm \frac{2 \sqrt{2}}{3}\)
AP EAMCET-2016
Hyperbola
120791
the equation of a tangent to the hyperbola \(16 x\) \(-25 y^2-96 x+100 y-356=0\) which makes an angle \(\mathbf{4 5}^{\circ}\) with its transverse axis is
120792
Let origin be the centre and \(( \pm 3,0)\) be the foci and \(\frac{3}{2}\) be the eccentricity of hyperbola. Then the line \(2 \mathbf{x}-\mathbf{y}-\mathbf{1}=\mathbf{0}\)
1 intersects the hyperbola at two points
2 does not intersect the hyperbola
3 touches the hyperbola
4 passes through the vertex of the hyperbola
Explanation:
B Given, centre ( 0,0\()\)
foci \(( \pm 3,0), \mathrm{e}=\frac{3}{2}\)
ae \(=3\).
\(\mathrm{a}=\frac{3}{\mathrm{e}}=3 \times \frac{2}{3}\)
Also,
\(a=2\)
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=4\left(\frac{9}{4}-1\right)\)
\(\mathrm{b}^2=5\)
Equation of hyperbola is -
\(\frac{x^2}{4}-\frac{y^2}{5}=1\)
Now, line: \(2 \mathrm{x}-\mathrm{y}-1=0\)
\(\mathrm{y}=2 \mathrm{x}-1 . \quad\) (put in above equation)
\(\frac{x^2}{4}-\frac{(2 x-1)^2}{5}=1\)
\(\frac{x^2}{4}-\frac{\left(4 x^2+1-4 x\right)}{5}=1\)
\(5 x^2-16 x^2-4+16 x=20\)
\(11 x^2-16 x+24=0\)
Discriminant, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0\)
\((-16)^2-4 \times 11 \times 24\lt 0\)
\(256-44 \times 24\lt 0\)Hence, line \(2 \mathrm{x}-\mathrm{y}-1=0\) does not intersect the hyperbola.
120789
The locus of the mid-points of the chords of the circle \(x^2+y^2=16\) which are the tangents to the hyperbola \(9 x^2-16 y^2=144\) is
1 \(3 x^2-4 y^2=16\left(x^2+y^2\right)\)
2 \(4 x^2-3 y^2=9\left(x^2+y^2\right)\)
3 \(16 x^2-9 y^2=\left(x^2+y^2\right)^2\)
4 \(16 x^2-9 y^2=4\left(x^2+y^2\right)\)
Explanation:
C Given,
\(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { be a point in the locus }\)
\(\mathrm{x}^2+\mathrm{y}^2=16 \text { having } \mathrm{P} \text { is its midpoint }\)
\(\mathrm{xx}_1+\mathrm{yy}_1-16=\mathrm{x}_1^2+\mathrm{y}_1^2-16\)
\(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\)
If (1) is a tangent to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(16\left(\mathrm{x}_1\right)^2-9\left(\mathrm{y}_1\right)^2=\left(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\right)^2\)\(\therefore\) The locus of P is \(16 \mathrm{x}^2-9 \mathrm{y}^2=\left(\mathrm{x}^2+\mathrm{y}^2\right)^2\)
AP EAMCET-21.04.2019
Hyperbola
120790
The values of \(\mathrm{m}\) for which the line \(\mathrm{y}=\mathrm{mx}+2\) becomes a tangent to the hyperbola \(4 \mathrm{x}^2-9 \mathrm{y}^2=\) 36 is
1 \(\pm \frac{2}{3}\)
2 \(\pm \frac{2 \sqrt{2}}{3}\)
3 \(\pm \frac{8}{9}\)
4 \(\pm \frac{4 \sqrt{2}}{3}\)
Explanation:
B Given, \(4 x^2-9 y^2=36\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
We know that for a tangent to the hyperbola,
\(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\((2)^2=9 \mathrm{~m}^2-4\)
\(\therefore \quad(2)^2=9 \mathrm{~m}^2-4 \quad\{\) from, \(\mathrm{y}=\mathrm{mx}+2\}\)
\(\Rightarrow \mathrm{m}= \pm \frac{2 \sqrt{2}}{3}\)
AP EAMCET-2016
Hyperbola
120791
the equation of a tangent to the hyperbola \(16 x\) \(-25 y^2-96 x+100 y-356=0\) which makes an angle \(\mathbf{4 5}^{\circ}\) with its transverse axis is
120792
Let origin be the centre and \(( \pm 3,0)\) be the foci and \(\frac{3}{2}\) be the eccentricity of hyperbola. Then the line \(2 \mathbf{x}-\mathbf{y}-\mathbf{1}=\mathbf{0}\)
1 intersects the hyperbola at two points
2 does not intersect the hyperbola
3 touches the hyperbola
4 passes through the vertex of the hyperbola
Explanation:
B Given, centre ( 0,0\()\)
foci \(( \pm 3,0), \mathrm{e}=\frac{3}{2}\)
ae \(=3\).
\(\mathrm{a}=\frac{3}{\mathrm{e}}=3 \times \frac{2}{3}\)
Also,
\(a=2\)
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=4\left(\frac{9}{4}-1\right)\)
\(\mathrm{b}^2=5\)
Equation of hyperbola is -
\(\frac{x^2}{4}-\frac{y^2}{5}=1\)
Now, line: \(2 \mathrm{x}-\mathrm{y}-1=0\)
\(\mathrm{y}=2 \mathrm{x}-1 . \quad\) (put in above equation)
\(\frac{x^2}{4}-\frac{(2 x-1)^2}{5}=1\)
\(\frac{x^2}{4}-\frac{\left(4 x^2+1-4 x\right)}{5}=1\)
\(5 x^2-16 x^2-4+16 x=20\)
\(11 x^2-16 x+24=0\)
Discriminant, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0\)
\((-16)^2-4 \times 11 \times 24\lt 0\)
\(256-44 \times 24\lt 0\)Hence, line \(2 \mathrm{x}-\mathrm{y}-1=0\) does not intersect the hyperbola.
120789
The locus of the mid-points of the chords of the circle \(x^2+y^2=16\) which are the tangents to the hyperbola \(9 x^2-16 y^2=144\) is
1 \(3 x^2-4 y^2=16\left(x^2+y^2\right)\)
2 \(4 x^2-3 y^2=9\left(x^2+y^2\right)\)
3 \(16 x^2-9 y^2=\left(x^2+y^2\right)^2\)
4 \(16 x^2-9 y^2=4\left(x^2+y^2\right)\)
Explanation:
C Given,
\(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { be a point in the locus }\)
\(\mathrm{x}^2+\mathrm{y}^2=16 \text { having } \mathrm{P} \text { is its midpoint }\)
\(\mathrm{xx}_1+\mathrm{yy}_1-16=\mathrm{x}_1^2+\mathrm{y}_1^2-16\)
\(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\)
If (1) is a tangent to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(16\left(\mathrm{x}_1\right)^2-9\left(\mathrm{y}_1\right)^2=\left(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\right)^2\)\(\therefore\) The locus of P is \(16 \mathrm{x}^2-9 \mathrm{y}^2=\left(\mathrm{x}^2+\mathrm{y}^2\right)^2\)
AP EAMCET-21.04.2019
Hyperbola
120790
The values of \(\mathrm{m}\) for which the line \(\mathrm{y}=\mathrm{mx}+2\) becomes a tangent to the hyperbola \(4 \mathrm{x}^2-9 \mathrm{y}^2=\) 36 is
1 \(\pm \frac{2}{3}\)
2 \(\pm \frac{2 \sqrt{2}}{3}\)
3 \(\pm \frac{8}{9}\)
4 \(\pm \frac{4 \sqrt{2}}{3}\)
Explanation:
B Given, \(4 x^2-9 y^2=36\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
We know that for a tangent to the hyperbola,
\(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\((2)^2=9 \mathrm{~m}^2-4\)
\(\therefore \quad(2)^2=9 \mathrm{~m}^2-4 \quad\{\) from, \(\mathrm{y}=\mathrm{mx}+2\}\)
\(\Rightarrow \mathrm{m}= \pm \frac{2 \sqrt{2}}{3}\)
AP EAMCET-2016
Hyperbola
120791
the equation of a tangent to the hyperbola \(16 x\) \(-25 y^2-96 x+100 y-356=0\) which makes an angle \(\mathbf{4 5}^{\circ}\) with its transverse axis is
120792
Let origin be the centre and \(( \pm 3,0)\) be the foci and \(\frac{3}{2}\) be the eccentricity of hyperbola. Then the line \(2 \mathbf{x}-\mathbf{y}-\mathbf{1}=\mathbf{0}\)
1 intersects the hyperbola at two points
2 does not intersect the hyperbola
3 touches the hyperbola
4 passes through the vertex of the hyperbola
Explanation:
B Given, centre ( 0,0\()\)
foci \(( \pm 3,0), \mathrm{e}=\frac{3}{2}\)
ae \(=3\).
\(\mathrm{a}=\frac{3}{\mathrm{e}}=3 \times \frac{2}{3}\)
Also,
\(a=2\)
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=4\left(\frac{9}{4}-1\right)\)
\(\mathrm{b}^2=5\)
Equation of hyperbola is -
\(\frac{x^2}{4}-\frac{y^2}{5}=1\)
Now, line: \(2 \mathrm{x}-\mathrm{y}-1=0\)
\(\mathrm{y}=2 \mathrm{x}-1 . \quad\) (put in above equation)
\(\frac{x^2}{4}-\frac{(2 x-1)^2}{5}=1\)
\(\frac{x^2}{4}-\frac{\left(4 x^2+1-4 x\right)}{5}=1\)
\(5 x^2-16 x^2-4+16 x=20\)
\(11 x^2-16 x+24=0\)
Discriminant, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0\)
\((-16)^2-4 \times 11 \times 24\lt 0\)
\(256-44 \times 24\lt 0\)Hence, line \(2 \mathrm{x}-\mathrm{y}-1=0\) does not intersect the hyperbola.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hyperbola
120789
The locus of the mid-points of the chords of the circle \(x^2+y^2=16\) which are the tangents to the hyperbola \(9 x^2-16 y^2=144\) is
1 \(3 x^2-4 y^2=16\left(x^2+y^2\right)\)
2 \(4 x^2-3 y^2=9\left(x^2+y^2\right)\)
3 \(16 x^2-9 y^2=\left(x^2+y^2\right)^2\)
4 \(16 x^2-9 y^2=4\left(x^2+y^2\right)\)
Explanation:
C Given,
\(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { be a point in the locus }\)
\(\mathrm{x}^2+\mathrm{y}^2=16 \text { having } \mathrm{P} \text { is its midpoint }\)
\(\mathrm{xx}_1+\mathrm{yy}_1-16=\mathrm{x}_1^2+\mathrm{y}_1^2-16\)
\(\mathrm{xx}_1+\mathrm{yy}_1=\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\)
If (1) is a tangent to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
\(16\left(\mathrm{x}_1\right)^2-9\left(\mathrm{y}_1\right)^2=\left(\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2\right)^2\)\(\therefore\) The locus of P is \(16 \mathrm{x}^2-9 \mathrm{y}^2=\left(\mathrm{x}^2+\mathrm{y}^2\right)^2\)
AP EAMCET-21.04.2019
Hyperbola
120790
The values of \(\mathrm{m}\) for which the line \(\mathrm{y}=\mathrm{mx}+2\) becomes a tangent to the hyperbola \(4 \mathrm{x}^2-9 \mathrm{y}^2=\) 36 is
1 \(\pm \frac{2}{3}\)
2 \(\pm \frac{2 \sqrt{2}}{3}\)
3 \(\pm \frac{8}{9}\)
4 \(\pm \frac{4 \sqrt{2}}{3}\)
Explanation:
B Given, \(4 x^2-9 y^2=36\)
\(\frac{x^2}{9}-\frac{y^2}{4}=1\)
We know that for a tangent to the hyperbola,
\(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\)
\((2)^2=9 \mathrm{~m}^2-4\)
\(\therefore \quad(2)^2=9 \mathrm{~m}^2-4 \quad\{\) from, \(\mathrm{y}=\mathrm{mx}+2\}\)
\(\Rightarrow \mathrm{m}= \pm \frac{2 \sqrt{2}}{3}\)
AP EAMCET-2016
Hyperbola
120791
the equation of a tangent to the hyperbola \(16 x\) \(-25 y^2-96 x+100 y-356=0\) which makes an angle \(\mathbf{4 5}^{\circ}\) with its transverse axis is
120792
Let origin be the centre and \(( \pm 3,0)\) be the foci and \(\frac{3}{2}\) be the eccentricity of hyperbola. Then the line \(2 \mathbf{x}-\mathbf{y}-\mathbf{1}=\mathbf{0}\)
1 intersects the hyperbola at two points
2 does not intersect the hyperbola
3 touches the hyperbola
4 passes through the vertex of the hyperbola
Explanation:
B Given, centre ( 0,0\()\)
foci \(( \pm 3,0), \mathrm{e}=\frac{3}{2}\)
ae \(=3\).
\(\mathrm{a}=\frac{3}{\mathrm{e}}=3 \times \frac{2}{3}\)
Also,
\(a=2\)
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=4\left(\frac{9}{4}-1\right)\)
\(\mathrm{b}^2=5\)
Equation of hyperbola is -
\(\frac{x^2}{4}-\frac{y^2}{5}=1\)
Now, line: \(2 \mathrm{x}-\mathrm{y}-1=0\)
\(\mathrm{y}=2 \mathrm{x}-1 . \quad\) (put in above equation)
\(\frac{x^2}{4}-\frac{(2 x-1)^2}{5}=1\)
\(\frac{x^2}{4}-\frac{\left(4 x^2+1-4 x\right)}{5}=1\)
\(5 x^2-16 x^2-4+16 x=20\)
\(11 x^2-16 x+24=0\)
Discriminant, \(\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0\)
\((-16)^2-4 \times 11 \times 24\lt 0\)
\(256-44 \times 24\lt 0\)Hence, line \(2 \mathrm{x}-\mathrm{y}-1=0\) does not intersect the hyperbola.