NEET Test Series from KOTA - 10 Papers In MS WORD
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Ellipse
120499
Area of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) is given by
1 \(25 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(4 \pi\) sq. units
4 \(5 \pi\) sq. units
Explanation:
B Given,
Equation of ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\therefore \quad \mathrm{a}=5\) and \(\mathrm{b}=4\)
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\frac{\mathrm{y}^2}{16}=1-\frac{\mathrm{x}^2}{25}\)
\(y^2=\frac{16}{25}\left(25-x^2\right)\)
\(y=\frac{4}{5} \sqrt{25-x^2}\)
Since, the ellipse is symmetrical about the axes.
\(\therefore \text { Required area }(\mathrm{A})=4 \int_0^5 \frac{4}{5} \sqrt{25-\mathrm{x}^2} \mathrm{dx}\)
\(=4 \times \frac{4}{5} \int_0^5 \sqrt{5^2-\mathrm{x}^2} \mathrm{dx}\)
\(=\frac{16}{5}\left[\frac{\mathrm{x}}{2} \sqrt{(5)^2-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_0^5\)
\(=\frac{16}{5}\left[0+\frac{25}{2} \cdot \sin ^{-1}\left(\frac{5}{5}\right)-0-0\right]\)
\(=\frac{16}{5}\left[\frac{25}{2} \cdot \sin ^{-1}(1)\right]=\frac{16}{5}\left[\frac{25}{2} \cdot \frac{\pi}{2}\right]=20 \pi \text { sq. unit. }\)
COMEDK-2012
Ellipse
120501
The eccentricity of an ellipse, with centre at the origin, is \(\frac{1}{2}\). If one directrix is \(x=4\), its equation is
1 \(3 x^2+4 y^2=1\)
2 \(3 x^2+4 y^2=12\)
3 \(4 x^2+3 y^2=1\)
4 \(4 \mathrm{x}^2+3 \mathrm{y}^2=12\)
Explanation:
B Given that,
\(\text { Directrix is } \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\text { Eccentricity }=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
\(\mathrm{~b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=4\left(1-\frac{1}{4}\right)=3\)
\(\text { The ellipse is } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{3}=1\)
\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
Ans: b
Exp:(B) Given that,
Directrix is \(\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
Eccentricity \(=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
The ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
COMEDK-2016
Ellipse
120502
If the eccentricity of the hyperbola \(x^2-y^2 \sec ^2 \theta=4\) is \(\sqrt{3}\) times the eccentricity of the ellipse \(x^2 \sec ^2 \theta+y^2=16\) then the value of \(\theta\) equals
120507
The eccentricity of the ellipse, which meets the straight line \(\frac{x}{7}+\frac{y}{2}=1\) on the axis of \(x\) and the straight line \(\frac{x}{3}-\frac{y}{5}=1\) on the axis of \(y\) and whose axes lie along the axes of coordinates, is
1 \(\frac{3 \sqrt{2}}{7}\)
2 \(\frac{2 \sqrt{6}}{7}\)
3 \(\frac{\sqrt{3}}{7}\)
4 None of these
Explanation:
B Given,
\(\frac{x}{7}+\frac{y}{2}=1\)
\(\frac{x}{3}-\frac{y}{5}=1\)
Let the equation of the ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
It is given that it passes through \((7,0)\) and \((0,-5)\).
Therefore, \(\mathrm{a}^2=49\) and \(\mathrm{b}^2=25\)
The eccentricity of the ellipse is given by,
\(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(25=49\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{49}}=\sqrt{\frac{24}{49}}=\frac{2 \sqrt{6}}{7}\)
120499
Area of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) is given by
1 \(25 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(4 \pi\) sq. units
4 \(5 \pi\) sq. units
Explanation:
B Given,
Equation of ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\therefore \quad \mathrm{a}=5\) and \(\mathrm{b}=4\)
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\frac{\mathrm{y}^2}{16}=1-\frac{\mathrm{x}^2}{25}\)
\(y^2=\frac{16}{25}\left(25-x^2\right)\)
\(y=\frac{4}{5} \sqrt{25-x^2}\)
Since, the ellipse is symmetrical about the axes.
\(\therefore \text { Required area }(\mathrm{A})=4 \int_0^5 \frac{4}{5} \sqrt{25-\mathrm{x}^2} \mathrm{dx}\)
\(=4 \times \frac{4}{5} \int_0^5 \sqrt{5^2-\mathrm{x}^2} \mathrm{dx}\)
\(=\frac{16}{5}\left[\frac{\mathrm{x}}{2} \sqrt{(5)^2-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_0^5\)
\(=\frac{16}{5}\left[0+\frac{25}{2} \cdot \sin ^{-1}\left(\frac{5}{5}\right)-0-0\right]\)
\(=\frac{16}{5}\left[\frac{25}{2} \cdot \sin ^{-1}(1)\right]=\frac{16}{5}\left[\frac{25}{2} \cdot \frac{\pi}{2}\right]=20 \pi \text { sq. unit. }\)
COMEDK-2012
Ellipse
120501
The eccentricity of an ellipse, with centre at the origin, is \(\frac{1}{2}\). If one directrix is \(x=4\), its equation is
1 \(3 x^2+4 y^2=1\)
2 \(3 x^2+4 y^2=12\)
3 \(4 x^2+3 y^2=1\)
4 \(4 \mathrm{x}^2+3 \mathrm{y}^2=12\)
Explanation:
B Given that,
\(\text { Directrix is } \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\text { Eccentricity }=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
\(\mathrm{~b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=4\left(1-\frac{1}{4}\right)=3\)
\(\text { The ellipse is } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{3}=1\)
\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
Ans: b
Exp:(B) Given that,
Directrix is \(\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
Eccentricity \(=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
The ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
COMEDK-2016
Ellipse
120502
If the eccentricity of the hyperbola \(x^2-y^2 \sec ^2 \theta=4\) is \(\sqrt{3}\) times the eccentricity of the ellipse \(x^2 \sec ^2 \theta+y^2=16\) then the value of \(\theta\) equals
120507
The eccentricity of the ellipse, which meets the straight line \(\frac{x}{7}+\frac{y}{2}=1\) on the axis of \(x\) and the straight line \(\frac{x}{3}-\frac{y}{5}=1\) on the axis of \(y\) and whose axes lie along the axes of coordinates, is
1 \(\frac{3 \sqrt{2}}{7}\)
2 \(\frac{2 \sqrt{6}}{7}\)
3 \(\frac{\sqrt{3}}{7}\)
4 None of these
Explanation:
B Given,
\(\frac{x}{7}+\frac{y}{2}=1\)
\(\frac{x}{3}-\frac{y}{5}=1\)
Let the equation of the ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
It is given that it passes through \((7,0)\) and \((0,-5)\).
Therefore, \(\mathrm{a}^2=49\) and \(\mathrm{b}^2=25\)
The eccentricity of the ellipse is given by,
\(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(25=49\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{49}}=\sqrt{\frac{24}{49}}=\frac{2 \sqrt{6}}{7}\)
120499
Area of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) is given by
1 \(25 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(4 \pi\) sq. units
4 \(5 \pi\) sq. units
Explanation:
B Given,
Equation of ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\therefore \quad \mathrm{a}=5\) and \(\mathrm{b}=4\)
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\frac{\mathrm{y}^2}{16}=1-\frac{\mathrm{x}^2}{25}\)
\(y^2=\frac{16}{25}\left(25-x^2\right)\)
\(y=\frac{4}{5} \sqrt{25-x^2}\)
Since, the ellipse is symmetrical about the axes.
\(\therefore \text { Required area }(\mathrm{A})=4 \int_0^5 \frac{4}{5} \sqrt{25-\mathrm{x}^2} \mathrm{dx}\)
\(=4 \times \frac{4}{5} \int_0^5 \sqrt{5^2-\mathrm{x}^2} \mathrm{dx}\)
\(=\frac{16}{5}\left[\frac{\mathrm{x}}{2} \sqrt{(5)^2-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_0^5\)
\(=\frac{16}{5}\left[0+\frac{25}{2} \cdot \sin ^{-1}\left(\frac{5}{5}\right)-0-0\right]\)
\(=\frac{16}{5}\left[\frac{25}{2} \cdot \sin ^{-1}(1)\right]=\frac{16}{5}\left[\frac{25}{2} \cdot \frac{\pi}{2}\right]=20 \pi \text { sq. unit. }\)
COMEDK-2012
Ellipse
120501
The eccentricity of an ellipse, with centre at the origin, is \(\frac{1}{2}\). If one directrix is \(x=4\), its equation is
1 \(3 x^2+4 y^2=1\)
2 \(3 x^2+4 y^2=12\)
3 \(4 x^2+3 y^2=1\)
4 \(4 \mathrm{x}^2+3 \mathrm{y}^2=12\)
Explanation:
B Given that,
\(\text { Directrix is } \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\text { Eccentricity }=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
\(\mathrm{~b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=4\left(1-\frac{1}{4}\right)=3\)
\(\text { The ellipse is } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{3}=1\)
\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
Ans: b
Exp:(B) Given that,
Directrix is \(\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
Eccentricity \(=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
The ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
COMEDK-2016
Ellipse
120502
If the eccentricity of the hyperbola \(x^2-y^2 \sec ^2 \theta=4\) is \(\sqrt{3}\) times the eccentricity of the ellipse \(x^2 \sec ^2 \theta+y^2=16\) then the value of \(\theta\) equals
120507
The eccentricity of the ellipse, which meets the straight line \(\frac{x}{7}+\frac{y}{2}=1\) on the axis of \(x\) and the straight line \(\frac{x}{3}-\frac{y}{5}=1\) on the axis of \(y\) and whose axes lie along the axes of coordinates, is
1 \(\frac{3 \sqrt{2}}{7}\)
2 \(\frac{2 \sqrt{6}}{7}\)
3 \(\frac{\sqrt{3}}{7}\)
4 None of these
Explanation:
B Given,
\(\frac{x}{7}+\frac{y}{2}=1\)
\(\frac{x}{3}-\frac{y}{5}=1\)
Let the equation of the ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
It is given that it passes through \((7,0)\) and \((0,-5)\).
Therefore, \(\mathrm{a}^2=49\) and \(\mathrm{b}^2=25\)
The eccentricity of the ellipse is given by,
\(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(25=49\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{49}}=\sqrt{\frac{24}{49}}=\frac{2 \sqrt{6}}{7}\)
120499
Area of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) is given by
1 \(25 \pi\) sq. units
2 \(20 \pi\) sq. units
3 \(4 \pi\) sq. units
4 \(5 \pi\) sq. units
Explanation:
B Given,
Equation of ellipse, \(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\therefore \quad \mathrm{a}=5\) and \(\mathrm{b}=4\)
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(\frac{\mathrm{y}^2}{16}=1-\frac{\mathrm{x}^2}{25}\)
\(y^2=\frac{16}{25}\left(25-x^2\right)\)
\(y=\frac{4}{5} \sqrt{25-x^2}\)
Since, the ellipse is symmetrical about the axes.
\(\therefore \text { Required area }(\mathrm{A})=4 \int_0^5 \frac{4}{5} \sqrt{25-\mathrm{x}^2} \mathrm{dx}\)
\(=4 \times \frac{4}{5} \int_0^5 \sqrt{5^2-\mathrm{x}^2} \mathrm{dx}\)
\(=\frac{16}{5}\left[\frac{\mathrm{x}}{2} \sqrt{(5)^2-\mathrm{x}^2}+\frac{25}{2} \sin ^{-1} \frac{\mathrm{x}}{5}\right]_0^5\)
\(=\frac{16}{5}\left[0+\frac{25}{2} \cdot \sin ^{-1}\left(\frac{5}{5}\right)-0-0\right]\)
\(=\frac{16}{5}\left[\frac{25}{2} \cdot \sin ^{-1}(1)\right]=\frac{16}{5}\left[\frac{25}{2} \cdot \frac{\pi}{2}\right]=20 \pi \text { sq. unit. }\)
COMEDK-2012
Ellipse
120501
The eccentricity of an ellipse, with centre at the origin, is \(\frac{1}{2}\). If one directrix is \(x=4\), its equation is
1 \(3 x^2+4 y^2=1\)
2 \(3 x^2+4 y^2=12\)
3 \(4 x^2+3 y^2=1\)
4 \(4 \mathrm{x}^2+3 \mathrm{y}^2=12\)
Explanation:
B Given that,
\(\text { Directrix is } \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
\(\text { Eccentricity }=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
\(\mathrm{~b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=4\left(1-\frac{1}{4}\right)=3\)
\(\text { The ellipse is } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{3}=1\)
\(3 \mathrm{x}^2+4 \mathrm{y}^2=12\)
Ans: b
Exp:(B) Given that,
Directrix is \(\mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4\)
Eccentricity \(=\frac{1}{2}\)
\(\therefore \mathrm{a}=4 \mathrm{e}=4 \times \frac{1}{2}=2\)
The ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
COMEDK-2016
Ellipse
120502
If the eccentricity of the hyperbola \(x^2-y^2 \sec ^2 \theta=4\) is \(\sqrt{3}\) times the eccentricity of the ellipse \(x^2 \sec ^2 \theta+y^2=16\) then the value of \(\theta\) equals
120507
The eccentricity of the ellipse, which meets the straight line \(\frac{x}{7}+\frac{y}{2}=1\) on the axis of \(x\) and the straight line \(\frac{x}{3}-\frac{y}{5}=1\) on the axis of \(y\) and whose axes lie along the axes of coordinates, is
1 \(\frac{3 \sqrt{2}}{7}\)
2 \(\frac{2 \sqrt{6}}{7}\)
3 \(\frac{\sqrt{3}}{7}\)
4 None of these
Explanation:
B Given,
\(\frac{x}{7}+\frac{y}{2}=1\)
\(\frac{x}{3}-\frac{y}{5}=1\)
Let the equation of the ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
It is given that it passes through \((7,0)\) and \((0,-5)\).
Therefore, \(\mathrm{a}^2=49\) and \(\mathrm{b}^2=25\)
The eccentricity of the ellipse is given by,
\(\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(25=49\left(1-\mathrm{e}^2\right)\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{25}{49}}=\sqrt{\frac{24}{49}}=\frac{2 \sqrt{6}}{7}\)
