120577
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is
1 \(\frac{1}{4}(\sqrt{5}-1)\)
2 \(\frac{1}{2}(\sqrt{5}+1)\)
3 \(\frac{1}{2}(\sqrt{5}-1)\)
4 \(\frac{1}{4}(\sqrt{5}+1)\)
Explanation:
C According to question,
The difference between foci = length of latus rectum
\(2 a e=\frac{2 b^2}{a}\)
\(e=\frac{b^2}{a^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}^2=1-\mathrm{e}\)
\(\mathrm{e}^2+\mathrm{e}-1=0\)
And
\(0\lt \mathrm{e}\lt 1\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)
\(\mathrm{e}=\frac{1}{2}(\sqrt{5}-1)\)
(only positive value)
WB JEE-2013
Ellipse
120578
The equation of the ellipse whose foci are \(( \pm, 2,0)\) and eccentricity \(\frac{1}{2}\) is:
1 \(\frac{x^2}{12}+\frac{y^2}{16}=1\)
2 \(\frac{x^2}{16}+\frac{y^2}{12}=1\)
3 \(\frac{x^2}{16}+\frac{y^2}{8}=1\)
4 None of these
Explanation:
B Given, \(\mathrm{e}=\frac{1}{2}\)
And foci \(( \pm \mathrm{ae}, 0)=( \pm 2,0)\)
\(\therefore \mathrm{a} \times \frac{1}{2}=2\)
\(\therefore \mathrm{a}=4\)
As we know that,
\(\mathrm{e}^2=\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{1}{4}=\frac{16-b^2}{16}\)
\(b^2=12\)
So, the equation of ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
Jamia Millia Islamia-2005
Ellipse
120579
\(x=4(1+\cos \theta)\) and \(y=3(1+\sin \theta)\) are the parametric equation of
1 \(\frac{(x-3)^2}{9}+\frac{(y-4)^2}{16}=1\)
2 \(\frac{(x+4)^2}{16}+\frac{(y-3)^2}{9}=1\)
3 \(\frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1\)
4 \(\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1\)
Explanation:
D Given,
\(x=4(1+\cos \theta)\)
\(y=3(1+\sin \theta)\)
From eqn. (i) \& (ii), we get-
\(\cos \theta=\frac{x}{4}-1\)
\(\sin \theta=\frac{y}{3}-1\)
Squaring and adding equation (iii) \& (iv)
\(\cos ^2 \theta+\sin ^2 \theta=\left[\frac{\mathrm{x}}{4}-1\right]^2+\left[\frac{\mathrm{y}}{3}-1\right]^2\)
\(1=\frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}\)
\(\therefore \frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}=1\)
Jamia Millia Islamia-2008
Ellipse
120580
The sum of major and minor axes lengths of an ellipse whose eccentricity is \(4 / 5\) and length of latus rectum is 14.4 is
1 24
2 32
3 64
4 48
Explanation:
C Given,
Length of major axis of an ellipse \(=2 \mathrm{a}\)
Length of minor axis of an ellipse \(=2 \mathrm{~b}\)
Eccentricity \((\mathrm{e})=\frac{4}{5}\)
And length of latus rectum \(=14.4\)
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=14.4\)
\(\mathrm{~b}^2=7.2 \mathrm{a}\)
\(\text { Also, } \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(7 \cdot 2 \mathrm{a}=\mathrm{a}^2\left[1-\left(\frac{4}{5}\right)^2\right]\)
\(7.2 \mathrm{a}=\mathrm{a}^2\left(\frac{9}{25}\right)\)
\(\mathrm{a}=0 \text { or } \mathrm{a}=20\)
\(\text { But } \mathrm{a} \neq 0 \quad \mathrm{a}=20\)
\(\text { Putting } \mathrm{a}=20 \text { in } \mathrm{b}^2=7.2 \mathrm{a}, \text { we get- }\)
\(\mathrm{b}=12\)
The sum of major and minor axis \(=2 \mathrm{a}+2 \mathrm{~b}\)
\(=2(20+12)\)
\(=2(32)\)
\(=64 \text { units. }\)
But \(\mathrm{a} \neq 0 \quad \mathrm{a}=20\)
120577
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is
1 \(\frac{1}{4}(\sqrt{5}-1)\)
2 \(\frac{1}{2}(\sqrt{5}+1)\)
3 \(\frac{1}{2}(\sqrt{5}-1)\)
4 \(\frac{1}{4}(\sqrt{5}+1)\)
Explanation:
C According to question,
The difference between foci = length of latus rectum
\(2 a e=\frac{2 b^2}{a}\)
\(e=\frac{b^2}{a^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}^2=1-\mathrm{e}\)
\(\mathrm{e}^2+\mathrm{e}-1=0\)
And
\(0\lt \mathrm{e}\lt 1\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)
\(\mathrm{e}=\frac{1}{2}(\sqrt{5}-1)\)
(only positive value)
WB JEE-2013
Ellipse
120578
The equation of the ellipse whose foci are \(( \pm, 2,0)\) and eccentricity \(\frac{1}{2}\) is:
1 \(\frac{x^2}{12}+\frac{y^2}{16}=1\)
2 \(\frac{x^2}{16}+\frac{y^2}{12}=1\)
3 \(\frac{x^2}{16}+\frac{y^2}{8}=1\)
4 None of these
Explanation:
B Given, \(\mathrm{e}=\frac{1}{2}\)
And foci \(( \pm \mathrm{ae}, 0)=( \pm 2,0)\)
\(\therefore \mathrm{a} \times \frac{1}{2}=2\)
\(\therefore \mathrm{a}=4\)
As we know that,
\(\mathrm{e}^2=\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{1}{4}=\frac{16-b^2}{16}\)
\(b^2=12\)
So, the equation of ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
Jamia Millia Islamia-2005
Ellipse
120579
\(x=4(1+\cos \theta)\) and \(y=3(1+\sin \theta)\) are the parametric equation of
1 \(\frac{(x-3)^2}{9}+\frac{(y-4)^2}{16}=1\)
2 \(\frac{(x+4)^2}{16}+\frac{(y-3)^2}{9}=1\)
3 \(\frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1\)
4 \(\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1\)
Explanation:
D Given,
\(x=4(1+\cos \theta)\)
\(y=3(1+\sin \theta)\)
From eqn. (i) \& (ii), we get-
\(\cos \theta=\frac{x}{4}-1\)
\(\sin \theta=\frac{y}{3}-1\)
Squaring and adding equation (iii) \& (iv)
\(\cos ^2 \theta+\sin ^2 \theta=\left[\frac{\mathrm{x}}{4}-1\right]^2+\left[\frac{\mathrm{y}}{3}-1\right]^2\)
\(1=\frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}\)
\(\therefore \frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}=1\)
Jamia Millia Islamia-2008
Ellipse
120580
The sum of major and minor axes lengths of an ellipse whose eccentricity is \(4 / 5\) and length of latus rectum is 14.4 is
1 24
2 32
3 64
4 48
Explanation:
C Given,
Length of major axis of an ellipse \(=2 \mathrm{a}\)
Length of minor axis of an ellipse \(=2 \mathrm{~b}\)
Eccentricity \((\mathrm{e})=\frac{4}{5}\)
And length of latus rectum \(=14.4\)
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=14.4\)
\(\mathrm{~b}^2=7.2 \mathrm{a}\)
\(\text { Also, } \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(7 \cdot 2 \mathrm{a}=\mathrm{a}^2\left[1-\left(\frac{4}{5}\right)^2\right]\)
\(7.2 \mathrm{a}=\mathrm{a}^2\left(\frac{9}{25}\right)\)
\(\mathrm{a}=0 \text { or } \mathrm{a}=20\)
\(\text { But } \mathrm{a} \neq 0 \quad \mathrm{a}=20\)
\(\text { Putting } \mathrm{a}=20 \text { in } \mathrm{b}^2=7.2 \mathrm{a}, \text { we get- }\)
\(\mathrm{b}=12\)
The sum of major and minor axis \(=2 \mathrm{a}+2 \mathrm{~b}\)
\(=2(20+12)\)
\(=2(32)\)
\(=64 \text { units. }\)
But \(\mathrm{a} \neq 0 \quad \mathrm{a}=20\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ellipse
120577
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is
1 \(\frac{1}{4}(\sqrt{5}-1)\)
2 \(\frac{1}{2}(\sqrt{5}+1)\)
3 \(\frac{1}{2}(\sqrt{5}-1)\)
4 \(\frac{1}{4}(\sqrt{5}+1)\)
Explanation:
C According to question,
The difference between foci = length of latus rectum
\(2 a e=\frac{2 b^2}{a}\)
\(e=\frac{b^2}{a^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}^2=1-\mathrm{e}\)
\(\mathrm{e}^2+\mathrm{e}-1=0\)
And
\(0\lt \mathrm{e}\lt 1\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)
\(\mathrm{e}=\frac{1}{2}(\sqrt{5}-1)\)
(only positive value)
WB JEE-2013
Ellipse
120578
The equation of the ellipse whose foci are \(( \pm, 2,0)\) and eccentricity \(\frac{1}{2}\) is:
1 \(\frac{x^2}{12}+\frac{y^2}{16}=1\)
2 \(\frac{x^2}{16}+\frac{y^2}{12}=1\)
3 \(\frac{x^2}{16}+\frac{y^2}{8}=1\)
4 None of these
Explanation:
B Given, \(\mathrm{e}=\frac{1}{2}\)
And foci \(( \pm \mathrm{ae}, 0)=( \pm 2,0)\)
\(\therefore \mathrm{a} \times \frac{1}{2}=2\)
\(\therefore \mathrm{a}=4\)
As we know that,
\(\mathrm{e}^2=\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{1}{4}=\frac{16-b^2}{16}\)
\(b^2=12\)
So, the equation of ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
Jamia Millia Islamia-2005
Ellipse
120579
\(x=4(1+\cos \theta)\) and \(y=3(1+\sin \theta)\) are the parametric equation of
1 \(\frac{(x-3)^2}{9}+\frac{(y-4)^2}{16}=1\)
2 \(\frac{(x+4)^2}{16}+\frac{(y-3)^2}{9}=1\)
3 \(\frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1\)
4 \(\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1\)
Explanation:
D Given,
\(x=4(1+\cos \theta)\)
\(y=3(1+\sin \theta)\)
From eqn. (i) \& (ii), we get-
\(\cos \theta=\frac{x}{4}-1\)
\(\sin \theta=\frac{y}{3}-1\)
Squaring and adding equation (iii) \& (iv)
\(\cos ^2 \theta+\sin ^2 \theta=\left[\frac{\mathrm{x}}{4}-1\right]^2+\left[\frac{\mathrm{y}}{3}-1\right]^2\)
\(1=\frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}\)
\(\therefore \frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}=1\)
Jamia Millia Islamia-2008
Ellipse
120580
The sum of major and minor axes lengths of an ellipse whose eccentricity is \(4 / 5\) and length of latus rectum is 14.4 is
1 24
2 32
3 64
4 48
Explanation:
C Given,
Length of major axis of an ellipse \(=2 \mathrm{a}\)
Length of minor axis of an ellipse \(=2 \mathrm{~b}\)
Eccentricity \((\mathrm{e})=\frac{4}{5}\)
And length of latus rectum \(=14.4\)
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=14.4\)
\(\mathrm{~b}^2=7.2 \mathrm{a}\)
\(\text { Also, } \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(7 \cdot 2 \mathrm{a}=\mathrm{a}^2\left[1-\left(\frac{4}{5}\right)^2\right]\)
\(7.2 \mathrm{a}=\mathrm{a}^2\left(\frac{9}{25}\right)\)
\(\mathrm{a}=0 \text { or } \mathrm{a}=20\)
\(\text { But } \mathrm{a} \neq 0 \quad \mathrm{a}=20\)
\(\text { Putting } \mathrm{a}=20 \text { in } \mathrm{b}^2=7.2 \mathrm{a}, \text { we get- }\)
\(\mathrm{b}=12\)
The sum of major and minor axis \(=2 \mathrm{a}+2 \mathrm{~b}\)
\(=2(20+12)\)
\(=2(32)\)
\(=64 \text { units. }\)
But \(\mathrm{a} \neq 0 \quad \mathrm{a}=20\)
120577
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is
1 \(\frac{1}{4}(\sqrt{5}-1)\)
2 \(\frac{1}{2}(\sqrt{5}+1)\)
3 \(\frac{1}{2}(\sqrt{5}-1)\)
4 \(\frac{1}{4}(\sqrt{5}+1)\)
Explanation:
C According to question,
The difference between foci = length of latus rectum
\(2 a e=\frac{2 b^2}{a}\)
\(e=\frac{b^2}{a^2}\)
Also, \(\quad \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}\)
\(\mathrm{e}=\sqrt{1-\mathrm{e}}\)
\(\mathrm{e}^2=1-\mathrm{e}\)
\(\mathrm{e}^2+\mathrm{e}-1=0\)
And
\(0\lt \mathrm{e}\lt 1\)
\(\mathrm{e}=\frac{-1 \pm \sqrt{1+4}}{2}\)
\(\mathrm{e}=\frac{1}{2}(\sqrt{5}-1)\)
(only positive value)
WB JEE-2013
Ellipse
120578
The equation of the ellipse whose foci are \(( \pm, 2,0)\) and eccentricity \(\frac{1}{2}\) is:
1 \(\frac{x^2}{12}+\frac{y^2}{16}=1\)
2 \(\frac{x^2}{16}+\frac{y^2}{12}=1\)
3 \(\frac{x^2}{16}+\frac{y^2}{8}=1\)
4 None of these
Explanation:
B Given, \(\mathrm{e}=\frac{1}{2}\)
And foci \(( \pm \mathrm{ae}, 0)=( \pm 2,0)\)
\(\therefore \mathrm{a} \times \frac{1}{2}=2\)
\(\therefore \mathrm{a}=4\)
As we know that,
\(\mathrm{e}^2=\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{a}^2}\)
\(\frac{1}{4}=\frac{16-b^2}{16}\)
\(b^2=12\)
So, the equation of ellipse
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{12}=1\)
Jamia Millia Islamia-2005
Ellipse
120579
\(x=4(1+\cos \theta)\) and \(y=3(1+\sin \theta)\) are the parametric equation of
1 \(\frac{(x-3)^2}{9}+\frac{(y-4)^2}{16}=1\)
2 \(\frac{(x+4)^2}{16}+\frac{(y-3)^2}{9}=1\)
3 \(\frac{(x-4)^2}{16}-\frac{(y-3)^2}{9}=1\)
4 \(\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1\)
Explanation:
D Given,
\(x=4(1+\cos \theta)\)
\(y=3(1+\sin \theta)\)
From eqn. (i) \& (ii), we get-
\(\cos \theta=\frac{x}{4}-1\)
\(\sin \theta=\frac{y}{3}-1\)
Squaring and adding equation (iii) \& (iv)
\(\cos ^2 \theta+\sin ^2 \theta=\left[\frac{\mathrm{x}}{4}-1\right]^2+\left[\frac{\mathrm{y}}{3}-1\right]^2\)
\(1=\frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}\)
\(\therefore \frac{(\mathrm{x}-4)^2}{16}+\frac{(\mathrm{y}-3)^2}{9}=1\)
Jamia Millia Islamia-2008
Ellipse
120580
The sum of major and minor axes lengths of an ellipse whose eccentricity is \(4 / 5\) and length of latus rectum is 14.4 is
1 24
2 32
3 64
4 48
Explanation:
C Given,
Length of major axis of an ellipse \(=2 \mathrm{a}\)
Length of minor axis of an ellipse \(=2 \mathrm{~b}\)
Eccentricity \((\mathrm{e})=\frac{4}{5}\)
And length of latus rectum \(=14.4\)
\(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=14.4\)
\(\mathrm{~b}^2=7.2 \mathrm{a}\)
\(\text { Also, } \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\)
\(7 \cdot 2 \mathrm{a}=\mathrm{a}^2\left[1-\left(\frac{4}{5}\right)^2\right]\)
\(7.2 \mathrm{a}=\mathrm{a}^2\left(\frac{9}{25}\right)\)
\(\mathrm{a}=0 \text { or } \mathrm{a}=20\)
\(\text { But } \mathrm{a} \neq 0 \quad \mathrm{a}=20\)
\(\text { Putting } \mathrm{a}=20 \text { in } \mathrm{b}^2=7.2 \mathrm{a}, \text { we get- }\)
\(\mathrm{b}=12\)
The sum of major and minor axis \(=2 \mathrm{a}+2 \mathrm{~b}\)
\(=2(20+12)\)
\(=2(32)\)
\(=64 \text { units. }\)
But \(\mathrm{a} \neq 0 \quad \mathrm{a}=20\)