NEET Test Series from KOTA - 10 Papers In MS WORD
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Ellipse
120513
The equation of the ellipse, whose focus is the point \((-1,1)\), whose directrix is the straight line \(x-y+3=0\) and whose eccentricity is \(\frac{1}{2}\), is
1 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+3)^2\)
2 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+1)^2\)
3 \((x+1)^2+(y-1)^2=\frac{1}{6}(x-y+3)^2\)
4 \((x+1)^2+(y-1)^2=\frac{1}{2}(x-y+3)^2\)
Explanation:
A Given that, for equation of the ellipse whose focus is \(S(-1,1)\) and directrix(M) is \(x-y+3=0\) and eccentricity (e) is given as \(\frac{1}{2}\).
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the ellipse.
We know that distance between the points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\) is,
\(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
We know that, the perpendicular distance from the point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the line \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) is
\(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\)
\(\mathrm{SP}=\mathrm{ePM}\)
\(\mathrm{SP}^2=\mathrm{e}^2 \mathrm{PM}^2\)
\(\mathrm{SP}^2=\frac{1}{4} \mathrm{PM}^2\)
\(4 \mathrm{SP}^2=\mathrm{PM}^2\)
\(4\left[(\mathrm{x}+1)^2+(\mathrm{y}-1)^2\right]=\left(\frac{\mathrm{x}-\mathrm{y}+3}{\sqrt{(1)^2+(-1)^2}}\right)^2\)
\((\mathrm{x}+1)^2+(\mathrm{y}-1)^2=\frac{1}{8}(\mathrm{x}-\mathrm{y}+3)^2\)
UPSEE-2017
Ellipse
120514
The eccentricity of an ellipse \(9 x^2+16 y^2=144\) is
1 \(\frac{\sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{\sqrt{7}}{4}\)
4 \(\frac{2}{5}\)
Explanation:
C Given equation of ellipse \(9 x^2+16 y^2=144\)
\(9 x^2+16 y^2=144\)
\(\frac{x^2}{\frac{144}{9}}+\frac{y^2}{\frac{144}{16}}=1\)
\(\frac{x^2}{16}+\frac{y^2}{9}=1\)
we know that, the general form of equation of the ellipse is,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
on comparing, we get -
\(\mathrm{a}=4, \quad \mathrm{~b}=3\)
\(\because \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4} .\)
UPSEE-2016
Ellipse
120515
In an ellipse, if the lines joining focus to the extremities of the minor axis form an equilateral triangle with the minor axis, then the eccentricity of the ellipse is
1 \(\frac{\sqrt{3}}{2}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{\frac{2}{3}}\)
Explanation:
A Let's consider, the horizontal ellipse \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), where \(\mathrm{a}^2>\mathrm{b}^2\) Since, \(\mathrm{PQR}\) is an equilateral triangle
\(\therefore \quad \mathrm{PQ}=\mathrm{QR}\)
\(\sqrt{(\mathrm{ae}-0)^2+(0-\mathrm{b})^2}=\sqrt{0+(2 \mathrm{~b})^2}\)
\(\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2=4 \mathrm{~b}^2\)
\(\mathrm{~b}^2=\frac{\mathrm{a}^2 \mathrm{e}^2}{3}\)
Now, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{\mathrm{e}^2}{3}}\)
\(\mathrm{e}^2=1-\frac{\mathrm{e}^2}{3}\)
\(\frac{4 \mathrm{e}^2}{3}=1\)
\(\mathrm{e}=\frac{\sqrt{3}}{2} .\)
UPSEE-2015
Ellipse
120516
The curve with parametric equations \(x=1+4 \cos \theta, y=2+3 \sin \theta i s\)
1 an ellipse
2 a parabola
3 a hyperbola
4 a circle
Explanation:
A Given that,
\(x=1+4 \cos \theta\)
\(y=2+3 \sin \theta\)
we have, \(x-1=4 \cos \theta, \cos \theta=\frac{x-1}{4}\)
\(\mathrm{y}-2=3 \sin \theta, \sin \theta=\frac{\mathrm{y}-2}{3}\)
we know that,
\(\sin ^2 \theta+\cos ^2 \theta=1\)
\(\left(\frac{x-1}{4}\right)^2+\left(\frac{y-2}{3}\right)^2=1\)
\(\frac{(x-1)^2}{16}+\frac{(y-2)^2}{9}=1\)Hence, which is an ellipse.
120513
The equation of the ellipse, whose focus is the point \((-1,1)\), whose directrix is the straight line \(x-y+3=0\) and whose eccentricity is \(\frac{1}{2}\), is
1 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+3)^2\)
2 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+1)^2\)
3 \((x+1)^2+(y-1)^2=\frac{1}{6}(x-y+3)^2\)
4 \((x+1)^2+(y-1)^2=\frac{1}{2}(x-y+3)^2\)
Explanation:
A Given that, for equation of the ellipse whose focus is \(S(-1,1)\) and directrix(M) is \(x-y+3=0\) and eccentricity (e) is given as \(\frac{1}{2}\).
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the ellipse.
We know that distance between the points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\) is,
\(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
We know that, the perpendicular distance from the point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the line \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) is
\(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\)
\(\mathrm{SP}=\mathrm{ePM}\)
\(\mathrm{SP}^2=\mathrm{e}^2 \mathrm{PM}^2\)
\(\mathrm{SP}^2=\frac{1}{4} \mathrm{PM}^2\)
\(4 \mathrm{SP}^2=\mathrm{PM}^2\)
\(4\left[(\mathrm{x}+1)^2+(\mathrm{y}-1)^2\right]=\left(\frac{\mathrm{x}-\mathrm{y}+3}{\sqrt{(1)^2+(-1)^2}}\right)^2\)
\((\mathrm{x}+1)^2+(\mathrm{y}-1)^2=\frac{1}{8}(\mathrm{x}-\mathrm{y}+3)^2\)
UPSEE-2017
Ellipse
120514
The eccentricity of an ellipse \(9 x^2+16 y^2=144\) is
1 \(\frac{\sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{\sqrt{7}}{4}\)
4 \(\frac{2}{5}\)
Explanation:
C Given equation of ellipse \(9 x^2+16 y^2=144\)
\(9 x^2+16 y^2=144\)
\(\frac{x^2}{\frac{144}{9}}+\frac{y^2}{\frac{144}{16}}=1\)
\(\frac{x^2}{16}+\frac{y^2}{9}=1\)
we know that, the general form of equation of the ellipse is,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
on comparing, we get -
\(\mathrm{a}=4, \quad \mathrm{~b}=3\)
\(\because \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4} .\)
UPSEE-2016
Ellipse
120515
In an ellipse, if the lines joining focus to the extremities of the minor axis form an equilateral triangle with the minor axis, then the eccentricity of the ellipse is
1 \(\frac{\sqrt{3}}{2}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{\frac{2}{3}}\)
Explanation:
A Let's consider, the horizontal ellipse \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), where \(\mathrm{a}^2>\mathrm{b}^2\) Since, \(\mathrm{PQR}\) is an equilateral triangle
\(\therefore \quad \mathrm{PQ}=\mathrm{QR}\)
\(\sqrt{(\mathrm{ae}-0)^2+(0-\mathrm{b})^2}=\sqrt{0+(2 \mathrm{~b})^2}\)
\(\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2=4 \mathrm{~b}^2\)
\(\mathrm{~b}^2=\frac{\mathrm{a}^2 \mathrm{e}^2}{3}\)
Now, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{\mathrm{e}^2}{3}}\)
\(\mathrm{e}^2=1-\frac{\mathrm{e}^2}{3}\)
\(\frac{4 \mathrm{e}^2}{3}=1\)
\(\mathrm{e}=\frac{\sqrt{3}}{2} .\)
UPSEE-2015
Ellipse
120516
The curve with parametric equations \(x=1+4 \cos \theta, y=2+3 \sin \theta i s\)
1 an ellipse
2 a parabola
3 a hyperbola
4 a circle
Explanation:
A Given that,
\(x=1+4 \cos \theta\)
\(y=2+3 \sin \theta\)
we have, \(x-1=4 \cos \theta, \cos \theta=\frac{x-1}{4}\)
\(\mathrm{y}-2=3 \sin \theta, \sin \theta=\frac{\mathrm{y}-2}{3}\)
we know that,
\(\sin ^2 \theta+\cos ^2 \theta=1\)
\(\left(\frac{x-1}{4}\right)^2+\left(\frac{y-2}{3}\right)^2=1\)
\(\frac{(x-1)^2}{16}+\frac{(y-2)^2}{9}=1\)Hence, which is an ellipse.
120513
The equation of the ellipse, whose focus is the point \((-1,1)\), whose directrix is the straight line \(x-y+3=0\) and whose eccentricity is \(\frac{1}{2}\), is
1 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+3)^2\)
2 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+1)^2\)
3 \((x+1)^2+(y-1)^2=\frac{1}{6}(x-y+3)^2\)
4 \((x+1)^2+(y-1)^2=\frac{1}{2}(x-y+3)^2\)
Explanation:
A Given that, for equation of the ellipse whose focus is \(S(-1,1)\) and directrix(M) is \(x-y+3=0\) and eccentricity (e) is given as \(\frac{1}{2}\).
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the ellipse.
We know that distance between the points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\) is,
\(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
We know that, the perpendicular distance from the point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the line \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) is
\(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\)
\(\mathrm{SP}=\mathrm{ePM}\)
\(\mathrm{SP}^2=\mathrm{e}^2 \mathrm{PM}^2\)
\(\mathrm{SP}^2=\frac{1}{4} \mathrm{PM}^2\)
\(4 \mathrm{SP}^2=\mathrm{PM}^2\)
\(4\left[(\mathrm{x}+1)^2+(\mathrm{y}-1)^2\right]=\left(\frac{\mathrm{x}-\mathrm{y}+3}{\sqrt{(1)^2+(-1)^2}}\right)^2\)
\((\mathrm{x}+1)^2+(\mathrm{y}-1)^2=\frac{1}{8}(\mathrm{x}-\mathrm{y}+3)^2\)
UPSEE-2017
Ellipse
120514
The eccentricity of an ellipse \(9 x^2+16 y^2=144\) is
1 \(\frac{\sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{\sqrt{7}}{4}\)
4 \(\frac{2}{5}\)
Explanation:
C Given equation of ellipse \(9 x^2+16 y^2=144\)
\(9 x^2+16 y^2=144\)
\(\frac{x^2}{\frac{144}{9}}+\frac{y^2}{\frac{144}{16}}=1\)
\(\frac{x^2}{16}+\frac{y^2}{9}=1\)
we know that, the general form of equation of the ellipse is,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
on comparing, we get -
\(\mathrm{a}=4, \quad \mathrm{~b}=3\)
\(\because \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4} .\)
UPSEE-2016
Ellipse
120515
In an ellipse, if the lines joining focus to the extremities of the minor axis form an equilateral triangle with the minor axis, then the eccentricity of the ellipse is
1 \(\frac{\sqrt{3}}{2}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{\frac{2}{3}}\)
Explanation:
A Let's consider, the horizontal ellipse \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), where \(\mathrm{a}^2>\mathrm{b}^2\) Since, \(\mathrm{PQR}\) is an equilateral triangle
\(\therefore \quad \mathrm{PQ}=\mathrm{QR}\)
\(\sqrt{(\mathrm{ae}-0)^2+(0-\mathrm{b})^2}=\sqrt{0+(2 \mathrm{~b})^2}\)
\(\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2=4 \mathrm{~b}^2\)
\(\mathrm{~b}^2=\frac{\mathrm{a}^2 \mathrm{e}^2}{3}\)
Now, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{\mathrm{e}^2}{3}}\)
\(\mathrm{e}^2=1-\frac{\mathrm{e}^2}{3}\)
\(\frac{4 \mathrm{e}^2}{3}=1\)
\(\mathrm{e}=\frac{\sqrt{3}}{2} .\)
UPSEE-2015
Ellipse
120516
The curve with parametric equations \(x=1+4 \cos \theta, y=2+3 \sin \theta i s\)
1 an ellipse
2 a parabola
3 a hyperbola
4 a circle
Explanation:
A Given that,
\(x=1+4 \cos \theta\)
\(y=2+3 \sin \theta\)
we have, \(x-1=4 \cos \theta, \cos \theta=\frac{x-1}{4}\)
\(\mathrm{y}-2=3 \sin \theta, \sin \theta=\frac{\mathrm{y}-2}{3}\)
we know that,
\(\sin ^2 \theta+\cos ^2 \theta=1\)
\(\left(\frac{x-1}{4}\right)^2+\left(\frac{y-2}{3}\right)^2=1\)
\(\frac{(x-1)^2}{16}+\frac{(y-2)^2}{9}=1\)Hence, which is an ellipse.
120513
The equation of the ellipse, whose focus is the point \((-1,1)\), whose directrix is the straight line \(x-y+3=0\) and whose eccentricity is \(\frac{1}{2}\), is
1 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+3)^2\)
2 \((x+1)^2+(y-1)^2=\frac{1}{8}(x-y+1)^2\)
3 \((x+1)^2+(y-1)^2=\frac{1}{6}(x-y+3)^2\)
4 \((x+1)^2+(y-1)^2=\frac{1}{2}(x-y+3)^2\)
Explanation:
A Given that, for equation of the ellipse whose focus is \(S(-1,1)\) and directrix(M) is \(x-y+3=0\) and eccentricity (e) is given as \(\frac{1}{2}\).
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on the ellipse.
We know that distance between the points \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) and \(\left(\mathrm{x}_2, \mathrm{y}_2\right)\) is,
\(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
We know that, the perpendicular distance from the point \(\left(\mathrm{x}_1, \mathrm{y}_1\right)\) to the line \(\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\) is
\(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\)
\(\mathrm{SP}=\mathrm{ePM}\)
\(\mathrm{SP}^2=\mathrm{e}^2 \mathrm{PM}^2\)
\(\mathrm{SP}^2=\frac{1}{4} \mathrm{PM}^2\)
\(4 \mathrm{SP}^2=\mathrm{PM}^2\)
\(4\left[(\mathrm{x}+1)^2+(\mathrm{y}-1)^2\right]=\left(\frac{\mathrm{x}-\mathrm{y}+3}{\sqrt{(1)^2+(-1)^2}}\right)^2\)
\((\mathrm{x}+1)^2+(\mathrm{y}-1)^2=\frac{1}{8}(\mathrm{x}-\mathrm{y}+3)^2\)
UPSEE-2017
Ellipse
120514
The eccentricity of an ellipse \(9 x^2+16 y^2=144\) is
1 \(\frac{\sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{\sqrt{7}}{4}\)
4 \(\frac{2}{5}\)
Explanation:
C Given equation of ellipse \(9 x^2+16 y^2=144\)
\(9 x^2+16 y^2=144\)
\(\frac{x^2}{\frac{144}{9}}+\frac{y^2}{\frac{144}{16}}=1\)
\(\frac{x^2}{16}+\frac{y^2}{9}=1\)
we know that, the general form of equation of the ellipse is,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
on comparing, we get -
\(\mathrm{a}=4, \quad \mathrm{~b}=3\)
\(\because \quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4} .\)
UPSEE-2016
Ellipse
120515
In an ellipse, if the lines joining focus to the extremities of the minor axis form an equilateral triangle with the minor axis, then the eccentricity of the ellipse is
1 \(\frac{\sqrt{3}}{2}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{\frac{2}{3}}\)
Explanation:
A Let's consider, the horizontal ellipse \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), where \(\mathrm{a}^2>\mathrm{b}^2\) Since, \(\mathrm{PQR}\) is an equilateral triangle
\(\therefore \quad \mathrm{PQ}=\mathrm{QR}\)
\(\sqrt{(\mathrm{ae}-0)^2+(0-\mathrm{b})^2}=\sqrt{0+(2 \mathrm{~b})^2}\)
\(\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2=4 \mathrm{~b}^2\)
\(\mathrm{~b}^2=\frac{\mathrm{a}^2 \mathrm{e}^2}{3}\)
Now, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{\mathrm{e}^2}{3}}\)
\(\mathrm{e}^2=1-\frac{\mathrm{e}^2}{3}\)
\(\frac{4 \mathrm{e}^2}{3}=1\)
\(\mathrm{e}=\frac{\sqrt{3}}{2} .\)
UPSEE-2015
Ellipse
120516
The curve with parametric equations \(x=1+4 \cos \theta, y=2+3 \sin \theta i s\)
1 an ellipse
2 a parabola
3 a hyperbola
4 a circle
Explanation:
A Given that,
\(x=1+4 \cos \theta\)
\(y=2+3 \sin \theta\)
we have, \(x-1=4 \cos \theta, \cos \theta=\frac{x-1}{4}\)
\(\mathrm{y}-2=3 \sin \theta, \sin \theta=\frac{\mathrm{y}-2}{3}\)
we know that,
\(\sin ^2 \theta+\cos ^2 \theta=1\)
\(\left(\frac{x-1}{4}\right)^2+\left(\frac{y-2}{3}\right)^2=1\)
\(\frac{(x-1)^2}{16}+\frac{(y-2)^2}{9}=1\)Hence, which is an ellipse.
