NEET Test Series from KOTA - 10 Papers In MS WORD
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Ellipse
120657
If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8 , then the eccentricity is
1 \(\frac{1}{\sqrt{5}}\)
2 \(\frac{1}{2}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
C The distance between the foci of an ellipse
\(2 \mathrm{ae}=6 \Rightarrow \mathrm{a}=\frac{3}{\mathrm{e}}\)
Length of the minor axis is \(2 b=8\)
\(b=4\)
\(b^2=16\)
We know that,
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{b}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=16\)
\(\frac{9}{\mathrm{e}^2}\left(1-\mathrm{e}^2\right)=16\)
\(9-9 \mathrm{e}^2=16 \mathrm{e}^2\)
\(25 \mathrm{e}^2=9\)
\(\mathrm{e}^2=\frac{9}{25}\)
\(\mathrm{e}=\frac{3}{5}\)
Manipal UGET-2019
Ellipse
120658
The number of real tangents that can be drawn to the ellipse \(3 x^2+5 y^2=32\) passing through (3, 5 ) is
1 0
2 1
3 2
4 infinite
Explanation:
C Given, the equation of the ellipse \(3 x^2+5 y^2=\) 32 or
\(\frac{\mathrm{x}^2}{\left(\frac{32}{3}\right)}+\frac{\mathrm{y}^2}{\left(\frac{32}{5}\right)}=1\)
\(\therefore \mathrm{a}^2=\frac{32}{3} \text { and } \mathrm{b}^2=\frac{32}{5}\)
The equation of tangent to the ellipse is
\(y=m x+\sqrt{a^2 m^2+b^2}\)
\(y=m x+\sqrt{\left(\frac{32}{3}\right) m^2+\left(\frac{32}{5}\right)}\)
Now, this line passes through \((3,5)\)
\(\therefore 5=3 \mathrm{~m}+\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})=\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})^2=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25+9 \mathrm{~m}^2-30 \mathrm{~m}=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25 \times 15+135 \mathrm{~m}^2-450 \mathrm{~m}=160 \mathrm{~m}^2+96\)
\(25 m^2+450 m-279=0\)
The above eqn. is quadratic in \(\mathrm{m}\) whose discriminant, \(\mathrm{D}>0\)
\(\because \mathrm{D}=(450)^2+4 \times 25 \times 279\)
\(\therefore\) There are two real values of \(\mathrm{m}\).
So, we get two tangents.
Manipal UGET-2010
Ellipse
120659
The foci of the conic section
\(25 x^2+16 y^2-150 x=175 \text { are }\)
1 \((0, \pm 3)\)
2 \((0, \pm 2)\)
3 \((3, \pm 3)\)
4 \((0, \pm 1)\)
Explanation:
C Given the conic section.
\(25 x^2+16 y^2-150 x=175\)
\(25\left(x^2-6 x\right)+16 y^2=175\)
\(25\left(x^2-6 x+9-9\right)+16 y^2=175\)
\(25\left[(x-3)^2\right]-225+16 y^2=175\)
\(25(x-3)^2+16 y^2=400\)
\(\frac{(x-3)^2}{16}+\frac{y^2}{25}=1\)
Here, \(\mathrm{a}^2=16, \mathrm{~b}^2=25\)
Now, \(1-\mathrm{e}^2=\frac{16}{25} \Rightarrow \mathrm{e}^2=1-\frac{16}{25}=\frac{9}{25}\)
\(\therefore \mathrm{e}= \pm \frac{3}{5}\)
\(\therefore \pm \mathrm{be}= \pm \frac{3}{5} \times 5= \pm 3\)Now, \(\mathrm{x}-3=0 \Rightarrow \mathrm{x}=3\)
\(\therefore\) The foci of the conic section are \((3, \pm 3)\).
Manipal UGET-2016
Ellipse
120660
Tangent to the ellipse \(\frac{x^2}{32}+\frac{y^2}{18}=1\) having slope \(\frac{-3}{4}\) meet the coordinate axis at \(A\) and \(B\). Then, the area of \(\triangle A O B\), where \(O\) is the origin, is
1 12 sq units
2 8 sq units
3 24 sq units
4 32 sq units
Explanation:
C Given, the ellipse
\(\frac{x^2}{32}+\frac{y^2}{18}=1\)
A line \(y=m x+c\) is a tangent to the ellipse when,
\(c^2=a^2 m^2+b^2\)
Here \(\mathrm{a}^2=32, \mathrm{~b}^2=18, \mathrm{~m}=\frac{-3}{4}\)
\(\therefore \quad \mathrm{c}^2=32 \times\left(\frac{9}{16}\right)+18=36\)
\(\therefore \quad c=6\)
\(\therefore\) The equation of tangent is
\(y=\frac{-3}{4} x+6\)
At, \(x\)-axis,
\(\mathrm{y}=0 \Rightarrow \frac{3}{4} \mathrm{x}=6\)
\(\therefore \quad \mathrm{x} =8\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0, \mathrm{y}=6\)
\(\therefore\) Area of \(\triangle \mathrm{OAB}\) is \(\frac{1}{2} \times 6 \times 8=24\) sq.unit
120657
If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8 , then the eccentricity is
1 \(\frac{1}{\sqrt{5}}\)
2 \(\frac{1}{2}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
C The distance between the foci of an ellipse
\(2 \mathrm{ae}=6 \Rightarrow \mathrm{a}=\frac{3}{\mathrm{e}}\)
Length of the minor axis is \(2 b=8\)
\(b=4\)
\(b^2=16\)
We know that,
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{b}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=16\)
\(\frac{9}{\mathrm{e}^2}\left(1-\mathrm{e}^2\right)=16\)
\(9-9 \mathrm{e}^2=16 \mathrm{e}^2\)
\(25 \mathrm{e}^2=9\)
\(\mathrm{e}^2=\frac{9}{25}\)
\(\mathrm{e}=\frac{3}{5}\)
Manipal UGET-2019
Ellipse
120658
The number of real tangents that can be drawn to the ellipse \(3 x^2+5 y^2=32\) passing through (3, 5 ) is
1 0
2 1
3 2
4 infinite
Explanation:
C Given, the equation of the ellipse \(3 x^2+5 y^2=\) 32 or
\(\frac{\mathrm{x}^2}{\left(\frac{32}{3}\right)}+\frac{\mathrm{y}^2}{\left(\frac{32}{5}\right)}=1\)
\(\therefore \mathrm{a}^2=\frac{32}{3} \text { and } \mathrm{b}^2=\frac{32}{5}\)
The equation of tangent to the ellipse is
\(y=m x+\sqrt{a^2 m^2+b^2}\)
\(y=m x+\sqrt{\left(\frac{32}{3}\right) m^2+\left(\frac{32}{5}\right)}\)
Now, this line passes through \((3,5)\)
\(\therefore 5=3 \mathrm{~m}+\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})=\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})^2=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25+9 \mathrm{~m}^2-30 \mathrm{~m}=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25 \times 15+135 \mathrm{~m}^2-450 \mathrm{~m}=160 \mathrm{~m}^2+96\)
\(25 m^2+450 m-279=0\)
The above eqn. is quadratic in \(\mathrm{m}\) whose discriminant, \(\mathrm{D}>0\)
\(\because \mathrm{D}=(450)^2+4 \times 25 \times 279\)
\(\therefore\) There are two real values of \(\mathrm{m}\).
So, we get two tangents.
Manipal UGET-2010
Ellipse
120659
The foci of the conic section
\(25 x^2+16 y^2-150 x=175 \text { are }\)
1 \((0, \pm 3)\)
2 \((0, \pm 2)\)
3 \((3, \pm 3)\)
4 \((0, \pm 1)\)
Explanation:
C Given the conic section.
\(25 x^2+16 y^2-150 x=175\)
\(25\left(x^2-6 x\right)+16 y^2=175\)
\(25\left(x^2-6 x+9-9\right)+16 y^2=175\)
\(25\left[(x-3)^2\right]-225+16 y^2=175\)
\(25(x-3)^2+16 y^2=400\)
\(\frac{(x-3)^2}{16}+\frac{y^2}{25}=1\)
Here, \(\mathrm{a}^2=16, \mathrm{~b}^2=25\)
Now, \(1-\mathrm{e}^2=\frac{16}{25} \Rightarrow \mathrm{e}^2=1-\frac{16}{25}=\frac{9}{25}\)
\(\therefore \mathrm{e}= \pm \frac{3}{5}\)
\(\therefore \pm \mathrm{be}= \pm \frac{3}{5} \times 5= \pm 3\)Now, \(\mathrm{x}-3=0 \Rightarrow \mathrm{x}=3\)
\(\therefore\) The foci of the conic section are \((3, \pm 3)\).
Manipal UGET-2016
Ellipse
120660
Tangent to the ellipse \(\frac{x^2}{32}+\frac{y^2}{18}=1\) having slope \(\frac{-3}{4}\) meet the coordinate axis at \(A\) and \(B\). Then, the area of \(\triangle A O B\), where \(O\) is the origin, is
1 12 sq units
2 8 sq units
3 24 sq units
4 32 sq units
Explanation:
C Given, the ellipse
\(\frac{x^2}{32}+\frac{y^2}{18}=1\)
A line \(y=m x+c\) is a tangent to the ellipse when,
\(c^2=a^2 m^2+b^2\)
Here \(\mathrm{a}^2=32, \mathrm{~b}^2=18, \mathrm{~m}=\frac{-3}{4}\)
\(\therefore \quad \mathrm{c}^2=32 \times\left(\frac{9}{16}\right)+18=36\)
\(\therefore \quad c=6\)
\(\therefore\) The equation of tangent is
\(y=\frac{-3}{4} x+6\)
At, \(x\)-axis,
\(\mathrm{y}=0 \Rightarrow \frac{3}{4} \mathrm{x}=6\)
\(\therefore \quad \mathrm{x} =8\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0, \mathrm{y}=6\)
\(\therefore\) Area of \(\triangle \mathrm{OAB}\) is \(\frac{1}{2} \times 6 \times 8=24\) sq.unit
120657
If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8 , then the eccentricity is
1 \(\frac{1}{\sqrt{5}}\)
2 \(\frac{1}{2}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
C The distance between the foci of an ellipse
\(2 \mathrm{ae}=6 \Rightarrow \mathrm{a}=\frac{3}{\mathrm{e}}\)
Length of the minor axis is \(2 b=8\)
\(b=4\)
\(b^2=16\)
We know that,
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{b}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=16\)
\(\frac{9}{\mathrm{e}^2}\left(1-\mathrm{e}^2\right)=16\)
\(9-9 \mathrm{e}^2=16 \mathrm{e}^2\)
\(25 \mathrm{e}^2=9\)
\(\mathrm{e}^2=\frac{9}{25}\)
\(\mathrm{e}=\frac{3}{5}\)
Manipal UGET-2019
Ellipse
120658
The number of real tangents that can be drawn to the ellipse \(3 x^2+5 y^2=32\) passing through (3, 5 ) is
1 0
2 1
3 2
4 infinite
Explanation:
C Given, the equation of the ellipse \(3 x^2+5 y^2=\) 32 or
\(\frac{\mathrm{x}^2}{\left(\frac{32}{3}\right)}+\frac{\mathrm{y}^2}{\left(\frac{32}{5}\right)}=1\)
\(\therefore \mathrm{a}^2=\frac{32}{3} \text { and } \mathrm{b}^2=\frac{32}{5}\)
The equation of tangent to the ellipse is
\(y=m x+\sqrt{a^2 m^2+b^2}\)
\(y=m x+\sqrt{\left(\frac{32}{3}\right) m^2+\left(\frac{32}{5}\right)}\)
Now, this line passes through \((3,5)\)
\(\therefore 5=3 \mathrm{~m}+\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})=\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})^2=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25+9 \mathrm{~m}^2-30 \mathrm{~m}=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25 \times 15+135 \mathrm{~m}^2-450 \mathrm{~m}=160 \mathrm{~m}^2+96\)
\(25 m^2+450 m-279=0\)
The above eqn. is quadratic in \(\mathrm{m}\) whose discriminant, \(\mathrm{D}>0\)
\(\because \mathrm{D}=(450)^2+4 \times 25 \times 279\)
\(\therefore\) There are two real values of \(\mathrm{m}\).
So, we get two tangents.
Manipal UGET-2010
Ellipse
120659
The foci of the conic section
\(25 x^2+16 y^2-150 x=175 \text { are }\)
1 \((0, \pm 3)\)
2 \((0, \pm 2)\)
3 \((3, \pm 3)\)
4 \((0, \pm 1)\)
Explanation:
C Given the conic section.
\(25 x^2+16 y^2-150 x=175\)
\(25\left(x^2-6 x\right)+16 y^2=175\)
\(25\left(x^2-6 x+9-9\right)+16 y^2=175\)
\(25\left[(x-3)^2\right]-225+16 y^2=175\)
\(25(x-3)^2+16 y^2=400\)
\(\frac{(x-3)^2}{16}+\frac{y^2}{25}=1\)
Here, \(\mathrm{a}^2=16, \mathrm{~b}^2=25\)
Now, \(1-\mathrm{e}^2=\frac{16}{25} \Rightarrow \mathrm{e}^2=1-\frac{16}{25}=\frac{9}{25}\)
\(\therefore \mathrm{e}= \pm \frac{3}{5}\)
\(\therefore \pm \mathrm{be}= \pm \frac{3}{5} \times 5= \pm 3\)Now, \(\mathrm{x}-3=0 \Rightarrow \mathrm{x}=3\)
\(\therefore\) The foci of the conic section are \((3, \pm 3)\).
Manipal UGET-2016
Ellipse
120660
Tangent to the ellipse \(\frac{x^2}{32}+\frac{y^2}{18}=1\) having slope \(\frac{-3}{4}\) meet the coordinate axis at \(A\) and \(B\). Then, the area of \(\triangle A O B\), where \(O\) is the origin, is
1 12 sq units
2 8 sq units
3 24 sq units
4 32 sq units
Explanation:
C Given, the ellipse
\(\frac{x^2}{32}+\frac{y^2}{18}=1\)
A line \(y=m x+c\) is a tangent to the ellipse when,
\(c^2=a^2 m^2+b^2\)
Here \(\mathrm{a}^2=32, \mathrm{~b}^2=18, \mathrm{~m}=\frac{-3}{4}\)
\(\therefore \quad \mathrm{c}^2=32 \times\left(\frac{9}{16}\right)+18=36\)
\(\therefore \quad c=6\)
\(\therefore\) The equation of tangent is
\(y=\frac{-3}{4} x+6\)
At, \(x\)-axis,
\(\mathrm{y}=0 \Rightarrow \frac{3}{4} \mathrm{x}=6\)
\(\therefore \quad \mathrm{x} =8\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0, \mathrm{y}=6\)
\(\therefore\) Area of \(\triangle \mathrm{OAB}\) is \(\frac{1}{2} \times 6 \times 8=24\) sq.unit
120657
If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8 , then the eccentricity is
1 \(\frac{1}{\sqrt{5}}\)
2 \(\frac{1}{2}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
C The distance between the foci of an ellipse
\(2 \mathrm{ae}=6 \Rightarrow \mathrm{a}=\frac{3}{\mathrm{e}}\)
Length of the minor axis is \(2 b=8\)
\(b=4\)
\(b^2=16\)
We know that,
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{b}^2\)
\(\mathrm{a}^2\left(1-\mathrm{e}^2\right)=16\)
\(\frac{9}{\mathrm{e}^2}\left(1-\mathrm{e}^2\right)=16\)
\(9-9 \mathrm{e}^2=16 \mathrm{e}^2\)
\(25 \mathrm{e}^2=9\)
\(\mathrm{e}^2=\frac{9}{25}\)
\(\mathrm{e}=\frac{3}{5}\)
Manipal UGET-2019
Ellipse
120658
The number of real tangents that can be drawn to the ellipse \(3 x^2+5 y^2=32\) passing through (3, 5 ) is
1 0
2 1
3 2
4 infinite
Explanation:
C Given, the equation of the ellipse \(3 x^2+5 y^2=\) 32 or
\(\frac{\mathrm{x}^2}{\left(\frac{32}{3}\right)}+\frac{\mathrm{y}^2}{\left(\frac{32}{5}\right)}=1\)
\(\therefore \mathrm{a}^2=\frac{32}{3} \text { and } \mathrm{b}^2=\frac{32}{5}\)
The equation of tangent to the ellipse is
\(y=m x+\sqrt{a^2 m^2+b^2}\)
\(y=m x+\sqrt{\left(\frac{32}{3}\right) m^2+\left(\frac{32}{5}\right)}\)
Now, this line passes through \((3,5)\)
\(\therefore 5=3 \mathrm{~m}+\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})=\sqrt{\left(\frac{32}{3}\right) \mathrm{m}^2+\frac{32}{5}}\)
\((5-3 \mathrm{~m})^2=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25+9 \mathrm{~m}^2-30 \mathrm{~m}=\frac{32}{3} \mathrm{~m}^2+\frac{32}{5}\)
\(25 \times 15+135 \mathrm{~m}^2-450 \mathrm{~m}=160 \mathrm{~m}^2+96\)
\(25 m^2+450 m-279=0\)
The above eqn. is quadratic in \(\mathrm{m}\) whose discriminant, \(\mathrm{D}>0\)
\(\because \mathrm{D}=(450)^2+4 \times 25 \times 279\)
\(\therefore\) There are two real values of \(\mathrm{m}\).
So, we get two tangents.
Manipal UGET-2010
Ellipse
120659
The foci of the conic section
\(25 x^2+16 y^2-150 x=175 \text { are }\)
1 \((0, \pm 3)\)
2 \((0, \pm 2)\)
3 \((3, \pm 3)\)
4 \((0, \pm 1)\)
Explanation:
C Given the conic section.
\(25 x^2+16 y^2-150 x=175\)
\(25\left(x^2-6 x\right)+16 y^2=175\)
\(25\left(x^2-6 x+9-9\right)+16 y^2=175\)
\(25\left[(x-3)^2\right]-225+16 y^2=175\)
\(25(x-3)^2+16 y^2=400\)
\(\frac{(x-3)^2}{16}+\frac{y^2}{25}=1\)
Here, \(\mathrm{a}^2=16, \mathrm{~b}^2=25\)
Now, \(1-\mathrm{e}^2=\frac{16}{25} \Rightarrow \mathrm{e}^2=1-\frac{16}{25}=\frac{9}{25}\)
\(\therefore \mathrm{e}= \pm \frac{3}{5}\)
\(\therefore \pm \mathrm{be}= \pm \frac{3}{5} \times 5= \pm 3\)Now, \(\mathrm{x}-3=0 \Rightarrow \mathrm{x}=3\)
\(\therefore\) The foci of the conic section are \((3, \pm 3)\).
Manipal UGET-2016
Ellipse
120660
Tangent to the ellipse \(\frac{x^2}{32}+\frac{y^2}{18}=1\) having slope \(\frac{-3}{4}\) meet the coordinate axis at \(A\) and \(B\). Then, the area of \(\triangle A O B\), where \(O\) is the origin, is
1 12 sq units
2 8 sq units
3 24 sq units
4 32 sq units
Explanation:
C Given, the ellipse
\(\frac{x^2}{32}+\frac{y^2}{18}=1\)
A line \(y=m x+c\) is a tangent to the ellipse when,
\(c^2=a^2 m^2+b^2\)
Here \(\mathrm{a}^2=32, \mathrm{~b}^2=18, \mathrm{~m}=\frac{-3}{4}\)
\(\therefore \quad \mathrm{c}^2=32 \times\left(\frac{9}{16}\right)+18=36\)
\(\therefore \quad c=6\)
\(\therefore\) The equation of tangent is
\(y=\frac{-3}{4} x+6\)
At, \(x\)-axis,
\(\mathrm{y}=0 \Rightarrow \frac{3}{4} \mathrm{x}=6\)
\(\therefore \quad \mathrm{x} =8\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0, \mathrm{y}=6\)
\(\therefore\) Area of \(\triangle \mathrm{OAB}\) is \(\frac{1}{2} \times 6 \times 8=24\) sq.unit
