120525
The sum of the focal distances of any point on the conic \(\frac{\mathbf{x}^2}{25}+\frac{y^2}{16}=1\) is
1 10
2 9
3 41
4 18
Explanation:
A : Given,
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(a^2=25\)
\(a=5\)
\(b^2=16\)
\(b=4\)
We know that, if \(\mathrm{P}\) is any point on the curve, then
Sum of focal distance \(=\) length of major axis.
\(\therefore \mathrm{S}_1 \mathrm{P}+\mathrm{S}_2 \mathrm{P}=2 \mathrm{a}=2 \times 5=10 \text {. }\)
BCECE-2010
Ellipse
120526
In an ellipse, if the lines joining focus to the extremities of the major 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 the equation of ellipse
\(\left(a^2>b^2\right)\)
Since, \(\mathrm{BB}\) 'S is an equilateral triangle
\(B S=B^{\prime} B\)
\(\sqrt{(a e-0)^2+(0-b)^2}=\sqrt{0+(2 b)^2}\)
\(a^2 e^2+b^2=(2 b)^2\)
\(3 b^2=a^2 e^2\)
\(b^2=\frac{a^2 e^2}{3}\)
Now,
\(b^2=a^2\left(1-e^2\right)\)
\(\frac{b^2}{a^2}=1-e^2\)
\(\frac{a^2 e^2}{3 a^2}=1-e^2\)
\(\frac{e^2}{3}=1-e^2\)
\(e^2\left(\frac{1}{3}+1\right)=1\)
\(e^2=\frac{3}{4}=\frac{\sqrt{3}}{2}\)
AMU-2009
Ellipse
120527
Let \(S\) and \(S^{\prime}\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on SS' as diameter intersects the ellipse in real and distinct points, then the eccentricity \(e\) of the ellipse satisfies
B Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, foci be \(S(a e, 0)\) and \(S(-a e, 0)\).
Equation of circle with SS' as diameters is -
\((x+a e)(x-a e)+(y+0)(y-0)=0\)
\(x^2+y^2=(a e)^2\)
\(y^2=(a e)^2-x^2\)
Putting value of equation (ii) in equation (i), we get -
\(\frac{x^2}{a^2}+\frac{(a e)^2-x^2}{b^2}=1\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-(a e)^2}{(b)^2}\)
\(x^2\left(\frac{b^2}{a^2}-1\right)=b^2-(a e)^2\)
\(x^2\left(-e^2\right)=a^2\left(1-2 e^2\right) \quad\left\{e=\sqrt{1-\frac{b^2}{a^2}}\right\}\)
\(x^2=\frac{a^2}{e^2}\left(2 e^2-1\right)\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
\(\because\) Circle intersects the ellipse on distinct points, so the value of \(\mathrm{x}\) will be real and distinct.
\(\therefore \quad 2 \mathrm{e}^2-1>0\)
\(\mathrm{e}^2>\frac{1}{2}\)
\(\therefore \mathrm{e} \in\left(\frac{1}{\sqrt{2}}, 1\right)\)
\(\mathrm{e}> \pm \frac{1}{\sqrt{2}}\)
AMU-2019
Ellipse
120528
If the ellipse \(\frac{x^2}{4}+y^2=1\) meets the ellipse \(x^2+\frac{y^2}{a^2}=1\) in four distinct points and \(a=b^2-5 b+7\), then \(b\) does not lie in
1 \([4,5]\)
2 \((-\infty, 2) \cup(3, \infty)\)
3 \((-\infty, 0)\)
4 \([2,3]\)
Explanation:
D For the two ellipses to intersect in four distinct points
\(a^2>1\)
\(a>1 \text { or } a\lt -1\)
For, \(\quad \mathrm{a}\lt -1\)
\(\mathrm{b}^2-5 \mathrm{~b}+7\lt -1\)
\(\mathrm{~b}^2-5 \mathrm{~b}+8\lt 0\)
\(\left(\mathrm{~b}-\frac{5}{2}\right)^2+\frac{7}{4}\lt 0 \text { (which is not possible) }\)
Now, for a \(>1\)
\(b^2-5 b+7>1\)
\(b^2-5 b+6>0\)
\(b \in(-\infty, 2) \cup(3, \infty)\)Hence, \(b\) does not lie on \([2,3]\)
AMU-2019
Ellipse
120529
The distance between the directrices of the ellipse us \(\frac{x^2}{4}+\frac{y^2}{9}=1\) is
1 \(\frac{9}{\sqrt{5}}\)
2 \(\frac{18}{\sqrt{5}}\)
3 \(\frac{24}{\sqrt{5}}\)
4 None of these
Explanation:
B Given, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Here, \(\quad a^2=9\)
\(\mathrm{b}^2=4\)
Eccentricity, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{4}{9}}\)
\(\mathrm{e}=\frac{\sqrt{5}}{3}\)
Distance between directrix is \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
\(=\frac{2 \times 3}{\frac{\sqrt{5}}{3}}\)
\(=\frac{18}{\sqrt{5}}\)
120525
The sum of the focal distances of any point on the conic \(\frac{\mathbf{x}^2}{25}+\frac{y^2}{16}=1\) is
1 10
2 9
3 41
4 18
Explanation:
A : Given,
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(a^2=25\)
\(a=5\)
\(b^2=16\)
\(b=4\)
We know that, if \(\mathrm{P}\) is any point on the curve, then
Sum of focal distance \(=\) length of major axis.
\(\therefore \mathrm{S}_1 \mathrm{P}+\mathrm{S}_2 \mathrm{P}=2 \mathrm{a}=2 \times 5=10 \text {. }\)
BCECE-2010
Ellipse
120526
In an ellipse, if the lines joining focus to the extremities of the major 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 the equation of ellipse
\(\left(a^2>b^2\right)\)
Since, \(\mathrm{BB}\) 'S is an equilateral triangle
\(B S=B^{\prime} B\)
\(\sqrt{(a e-0)^2+(0-b)^2}=\sqrt{0+(2 b)^2}\)
\(a^2 e^2+b^2=(2 b)^2\)
\(3 b^2=a^2 e^2\)
\(b^2=\frac{a^2 e^2}{3}\)
Now,
\(b^2=a^2\left(1-e^2\right)\)
\(\frac{b^2}{a^2}=1-e^2\)
\(\frac{a^2 e^2}{3 a^2}=1-e^2\)
\(\frac{e^2}{3}=1-e^2\)
\(e^2\left(\frac{1}{3}+1\right)=1\)
\(e^2=\frac{3}{4}=\frac{\sqrt{3}}{2}\)
AMU-2009
Ellipse
120527
Let \(S\) and \(S^{\prime}\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on SS' as diameter intersects the ellipse in real and distinct points, then the eccentricity \(e\) of the ellipse satisfies
B Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, foci be \(S(a e, 0)\) and \(S(-a e, 0)\).
Equation of circle with SS' as diameters is -
\((x+a e)(x-a e)+(y+0)(y-0)=0\)
\(x^2+y^2=(a e)^2\)
\(y^2=(a e)^2-x^2\)
Putting value of equation (ii) in equation (i), we get -
\(\frac{x^2}{a^2}+\frac{(a e)^2-x^2}{b^2}=1\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-(a e)^2}{(b)^2}\)
\(x^2\left(\frac{b^2}{a^2}-1\right)=b^2-(a e)^2\)
\(x^2\left(-e^2\right)=a^2\left(1-2 e^2\right) \quad\left\{e=\sqrt{1-\frac{b^2}{a^2}}\right\}\)
\(x^2=\frac{a^2}{e^2}\left(2 e^2-1\right)\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
\(\because\) Circle intersects the ellipse on distinct points, so the value of \(\mathrm{x}\) will be real and distinct.
\(\therefore \quad 2 \mathrm{e}^2-1>0\)
\(\mathrm{e}^2>\frac{1}{2}\)
\(\therefore \mathrm{e} \in\left(\frac{1}{\sqrt{2}}, 1\right)\)
\(\mathrm{e}> \pm \frac{1}{\sqrt{2}}\)
AMU-2019
Ellipse
120528
If the ellipse \(\frac{x^2}{4}+y^2=1\) meets the ellipse \(x^2+\frac{y^2}{a^2}=1\) in four distinct points and \(a=b^2-5 b+7\), then \(b\) does not lie in
1 \([4,5]\)
2 \((-\infty, 2) \cup(3, \infty)\)
3 \((-\infty, 0)\)
4 \([2,3]\)
Explanation:
D For the two ellipses to intersect in four distinct points
\(a^2>1\)
\(a>1 \text { or } a\lt -1\)
For, \(\quad \mathrm{a}\lt -1\)
\(\mathrm{b}^2-5 \mathrm{~b}+7\lt -1\)
\(\mathrm{~b}^2-5 \mathrm{~b}+8\lt 0\)
\(\left(\mathrm{~b}-\frac{5}{2}\right)^2+\frac{7}{4}\lt 0 \text { (which is not possible) }\)
Now, for a \(>1\)
\(b^2-5 b+7>1\)
\(b^2-5 b+6>0\)
\(b \in(-\infty, 2) \cup(3, \infty)\)Hence, \(b\) does not lie on \([2,3]\)
AMU-2019
Ellipse
120529
The distance between the directrices of the ellipse us \(\frac{x^2}{4}+\frac{y^2}{9}=1\) is
1 \(\frac{9}{\sqrt{5}}\)
2 \(\frac{18}{\sqrt{5}}\)
3 \(\frac{24}{\sqrt{5}}\)
4 None of these
Explanation:
B Given, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Here, \(\quad a^2=9\)
\(\mathrm{b}^2=4\)
Eccentricity, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{4}{9}}\)
\(\mathrm{e}=\frac{\sqrt{5}}{3}\)
Distance between directrix is \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
\(=\frac{2 \times 3}{\frac{\sqrt{5}}{3}}\)
\(=\frac{18}{\sqrt{5}}\)
120525
The sum of the focal distances of any point on the conic \(\frac{\mathbf{x}^2}{25}+\frac{y^2}{16}=1\) is
1 10
2 9
3 41
4 18
Explanation:
A : Given,
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(a^2=25\)
\(a=5\)
\(b^2=16\)
\(b=4\)
We know that, if \(\mathrm{P}\) is any point on the curve, then
Sum of focal distance \(=\) length of major axis.
\(\therefore \mathrm{S}_1 \mathrm{P}+\mathrm{S}_2 \mathrm{P}=2 \mathrm{a}=2 \times 5=10 \text {. }\)
BCECE-2010
Ellipse
120526
In an ellipse, if the lines joining focus to the extremities of the major 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 the equation of ellipse
\(\left(a^2>b^2\right)\)
Since, \(\mathrm{BB}\) 'S is an equilateral triangle
\(B S=B^{\prime} B\)
\(\sqrt{(a e-0)^2+(0-b)^2}=\sqrt{0+(2 b)^2}\)
\(a^2 e^2+b^2=(2 b)^2\)
\(3 b^2=a^2 e^2\)
\(b^2=\frac{a^2 e^2}{3}\)
Now,
\(b^2=a^2\left(1-e^2\right)\)
\(\frac{b^2}{a^2}=1-e^2\)
\(\frac{a^2 e^2}{3 a^2}=1-e^2\)
\(\frac{e^2}{3}=1-e^2\)
\(e^2\left(\frac{1}{3}+1\right)=1\)
\(e^2=\frac{3}{4}=\frac{\sqrt{3}}{2}\)
AMU-2009
Ellipse
120527
Let \(S\) and \(S^{\prime}\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on SS' as diameter intersects the ellipse in real and distinct points, then the eccentricity \(e\) of the ellipse satisfies
B Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, foci be \(S(a e, 0)\) and \(S(-a e, 0)\).
Equation of circle with SS' as diameters is -
\((x+a e)(x-a e)+(y+0)(y-0)=0\)
\(x^2+y^2=(a e)^2\)
\(y^2=(a e)^2-x^2\)
Putting value of equation (ii) in equation (i), we get -
\(\frac{x^2}{a^2}+\frac{(a e)^2-x^2}{b^2}=1\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-(a e)^2}{(b)^2}\)
\(x^2\left(\frac{b^2}{a^2}-1\right)=b^2-(a e)^2\)
\(x^2\left(-e^2\right)=a^2\left(1-2 e^2\right) \quad\left\{e=\sqrt{1-\frac{b^2}{a^2}}\right\}\)
\(x^2=\frac{a^2}{e^2}\left(2 e^2-1\right)\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
\(\because\) Circle intersects the ellipse on distinct points, so the value of \(\mathrm{x}\) will be real and distinct.
\(\therefore \quad 2 \mathrm{e}^2-1>0\)
\(\mathrm{e}^2>\frac{1}{2}\)
\(\therefore \mathrm{e} \in\left(\frac{1}{\sqrt{2}}, 1\right)\)
\(\mathrm{e}> \pm \frac{1}{\sqrt{2}}\)
AMU-2019
Ellipse
120528
If the ellipse \(\frac{x^2}{4}+y^2=1\) meets the ellipse \(x^2+\frac{y^2}{a^2}=1\) in four distinct points and \(a=b^2-5 b+7\), then \(b\) does not lie in
1 \([4,5]\)
2 \((-\infty, 2) \cup(3, \infty)\)
3 \((-\infty, 0)\)
4 \([2,3]\)
Explanation:
D For the two ellipses to intersect in four distinct points
\(a^2>1\)
\(a>1 \text { or } a\lt -1\)
For, \(\quad \mathrm{a}\lt -1\)
\(\mathrm{b}^2-5 \mathrm{~b}+7\lt -1\)
\(\mathrm{~b}^2-5 \mathrm{~b}+8\lt 0\)
\(\left(\mathrm{~b}-\frac{5}{2}\right)^2+\frac{7}{4}\lt 0 \text { (which is not possible) }\)
Now, for a \(>1\)
\(b^2-5 b+7>1\)
\(b^2-5 b+6>0\)
\(b \in(-\infty, 2) \cup(3, \infty)\)Hence, \(b\) does not lie on \([2,3]\)
AMU-2019
Ellipse
120529
The distance between the directrices of the ellipse us \(\frac{x^2}{4}+\frac{y^2}{9}=1\) is
1 \(\frac{9}{\sqrt{5}}\)
2 \(\frac{18}{\sqrt{5}}\)
3 \(\frac{24}{\sqrt{5}}\)
4 None of these
Explanation:
B Given, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Here, \(\quad a^2=9\)
\(\mathrm{b}^2=4\)
Eccentricity, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{4}{9}}\)
\(\mathrm{e}=\frac{\sqrt{5}}{3}\)
Distance between directrix is \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
\(=\frac{2 \times 3}{\frac{\sqrt{5}}{3}}\)
\(=\frac{18}{\sqrt{5}}\)
120525
The sum of the focal distances of any point on the conic \(\frac{\mathbf{x}^2}{25}+\frac{y^2}{16}=1\) is
1 10
2 9
3 41
4 18
Explanation:
A : Given,
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(a^2=25\)
\(a=5\)
\(b^2=16\)
\(b=4\)
We know that, if \(\mathrm{P}\) is any point on the curve, then
Sum of focal distance \(=\) length of major axis.
\(\therefore \mathrm{S}_1 \mathrm{P}+\mathrm{S}_2 \mathrm{P}=2 \mathrm{a}=2 \times 5=10 \text {. }\)
BCECE-2010
Ellipse
120526
In an ellipse, if the lines joining focus to the extremities of the major 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 the equation of ellipse
\(\left(a^2>b^2\right)\)
Since, \(\mathrm{BB}\) 'S is an equilateral triangle
\(B S=B^{\prime} B\)
\(\sqrt{(a e-0)^2+(0-b)^2}=\sqrt{0+(2 b)^2}\)
\(a^2 e^2+b^2=(2 b)^2\)
\(3 b^2=a^2 e^2\)
\(b^2=\frac{a^2 e^2}{3}\)
Now,
\(b^2=a^2\left(1-e^2\right)\)
\(\frac{b^2}{a^2}=1-e^2\)
\(\frac{a^2 e^2}{3 a^2}=1-e^2\)
\(\frac{e^2}{3}=1-e^2\)
\(e^2\left(\frac{1}{3}+1\right)=1\)
\(e^2=\frac{3}{4}=\frac{\sqrt{3}}{2}\)
AMU-2009
Ellipse
120527
Let \(S\) and \(S^{\prime}\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on SS' as diameter intersects the ellipse in real and distinct points, then the eccentricity \(e\) of the ellipse satisfies
B Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, foci be \(S(a e, 0)\) and \(S(-a e, 0)\).
Equation of circle with SS' as diameters is -
\((x+a e)(x-a e)+(y+0)(y-0)=0\)
\(x^2+y^2=(a e)^2\)
\(y^2=(a e)^2-x^2\)
Putting value of equation (ii) in equation (i), we get -
\(\frac{x^2}{a^2}+\frac{(a e)^2-x^2}{b^2}=1\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-(a e)^2}{(b)^2}\)
\(x^2\left(\frac{b^2}{a^2}-1\right)=b^2-(a e)^2\)
\(x^2\left(-e^2\right)=a^2\left(1-2 e^2\right) \quad\left\{e=\sqrt{1-\frac{b^2}{a^2}}\right\}\)
\(x^2=\frac{a^2}{e^2}\left(2 e^2-1\right)\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
\(\because\) Circle intersects the ellipse on distinct points, so the value of \(\mathrm{x}\) will be real and distinct.
\(\therefore \quad 2 \mathrm{e}^2-1>0\)
\(\mathrm{e}^2>\frac{1}{2}\)
\(\therefore \mathrm{e} \in\left(\frac{1}{\sqrt{2}}, 1\right)\)
\(\mathrm{e}> \pm \frac{1}{\sqrt{2}}\)
AMU-2019
Ellipse
120528
If the ellipse \(\frac{x^2}{4}+y^2=1\) meets the ellipse \(x^2+\frac{y^2}{a^2}=1\) in four distinct points and \(a=b^2-5 b+7\), then \(b\) does not lie in
1 \([4,5]\)
2 \((-\infty, 2) \cup(3, \infty)\)
3 \((-\infty, 0)\)
4 \([2,3]\)
Explanation:
D For the two ellipses to intersect in four distinct points
\(a^2>1\)
\(a>1 \text { or } a\lt -1\)
For, \(\quad \mathrm{a}\lt -1\)
\(\mathrm{b}^2-5 \mathrm{~b}+7\lt -1\)
\(\mathrm{~b}^2-5 \mathrm{~b}+8\lt 0\)
\(\left(\mathrm{~b}-\frac{5}{2}\right)^2+\frac{7}{4}\lt 0 \text { (which is not possible) }\)
Now, for a \(>1\)
\(b^2-5 b+7>1\)
\(b^2-5 b+6>0\)
\(b \in(-\infty, 2) \cup(3, \infty)\)Hence, \(b\) does not lie on \([2,3]\)
AMU-2019
Ellipse
120529
The distance between the directrices of the ellipse us \(\frac{x^2}{4}+\frac{y^2}{9}=1\) is
1 \(\frac{9}{\sqrt{5}}\)
2 \(\frac{18}{\sqrt{5}}\)
3 \(\frac{24}{\sqrt{5}}\)
4 None of these
Explanation:
B Given, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Here, \(\quad a^2=9\)
\(\mathrm{b}^2=4\)
Eccentricity, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{4}{9}}\)
\(\mathrm{e}=\frac{\sqrt{5}}{3}\)
Distance between directrix is \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
\(=\frac{2 \times 3}{\frac{\sqrt{5}}{3}}\)
\(=\frac{18}{\sqrt{5}}\)
120525
The sum of the focal distances of any point on the conic \(\frac{\mathbf{x}^2}{25}+\frac{y^2}{16}=1\) is
1 10
2 9
3 41
4 18
Explanation:
A : Given,
\(\frac{x^2}{25}+\frac{y^2}{16}=1\)
\(a^2=25\)
\(a=5\)
\(b^2=16\)
\(b=4\)
We know that, if \(\mathrm{P}\) is any point on the curve, then
Sum of focal distance \(=\) length of major axis.
\(\therefore \mathrm{S}_1 \mathrm{P}+\mathrm{S}_2 \mathrm{P}=2 \mathrm{a}=2 \times 5=10 \text {. }\)
BCECE-2010
Ellipse
120526
In an ellipse, if the lines joining focus to the extremities of the major 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 the equation of ellipse
\(\left(a^2>b^2\right)\)
Since, \(\mathrm{BB}\) 'S is an equilateral triangle
\(B S=B^{\prime} B\)
\(\sqrt{(a e-0)^2+(0-b)^2}=\sqrt{0+(2 b)^2}\)
\(a^2 e^2+b^2=(2 b)^2\)
\(3 b^2=a^2 e^2\)
\(b^2=\frac{a^2 e^2}{3}\)
Now,
\(b^2=a^2\left(1-e^2\right)\)
\(\frac{b^2}{a^2}=1-e^2\)
\(\frac{a^2 e^2}{3 a^2}=1-e^2\)
\(\frac{e^2}{3}=1-e^2\)
\(e^2\left(\frac{1}{3}+1\right)=1\)
\(e^2=\frac{3}{4}=\frac{\sqrt{3}}{2}\)
AMU-2009
Ellipse
120527
Let \(S\) and \(S^{\prime}\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on SS' as diameter intersects the ellipse in real and distinct points, then the eccentricity \(e\) of the ellipse satisfies
B Given,
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Let, foci be \(S(a e, 0)\) and \(S(-a e, 0)\).
Equation of circle with SS' as diameters is -
\((x+a e)(x-a e)+(y+0)(y-0)=0\)
\(x^2+y^2=(a e)^2\)
\(y^2=(a e)^2-x^2\)
Putting value of equation (ii) in equation (i), we get -
\(\frac{x^2}{a^2}+\frac{(a e)^2-x^2}{b^2}=1\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-(a e)^2}{(b)^2}\)
\(x^2\left(\frac{b^2}{a^2}-1\right)=b^2-(a e)^2\)
\(x^2\left(-e^2\right)=a^2\left(1-2 e^2\right) \quad\left\{e=\sqrt{1-\frac{b^2}{a^2}}\right\}\)
\(x^2=\frac{a^2}{e^2}\left(2 e^2-1\right)\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
\(\because\) Circle intersects the ellipse on distinct points, so the value of \(\mathrm{x}\) will be real and distinct.
\(\therefore \quad 2 \mathrm{e}^2-1>0\)
\(\mathrm{e}^2>\frac{1}{2}\)
\(\therefore \mathrm{e} \in\left(\frac{1}{\sqrt{2}}, 1\right)\)
\(\mathrm{e}> \pm \frac{1}{\sqrt{2}}\)
AMU-2019
Ellipse
120528
If the ellipse \(\frac{x^2}{4}+y^2=1\) meets the ellipse \(x^2+\frac{y^2}{a^2}=1\) in four distinct points and \(a=b^2-5 b+7\), then \(b\) does not lie in
1 \([4,5]\)
2 \((-\infty, 2) \cup(3, \infty)\)
3 \((-\infty, 0)\)
4 \([2,3]\)
Explanation:
D For the two ellipses to intersect in four distinct points
\(a^2>1\)
\(a>1 \text { or } a\lt -1\)
For, \(\quad \mathrm{a}\lt -1\)
\(\mathrm{b}^2-5 \mathrm{~b}+7\lt -1\)
\(\mathrm{~b}^2-5 \mathrm{~b}+8\lt 0\)
\(\left(\mathrm{~b}-\frac{5}{2}\right)^2+\frac{7}{4}\lt 0 \text { (which is not possible) }\)
Now, for a \(>1\)
\(b^2-5 b+7>1\)
\(b^2-5 b+6>0\)
\(b \in(-\infty, 2) \cup(3, \infty)\)Hence, \(b\) does not lie on \([2,3]\)
AMU-2019
Ellipse
120529
The distance between the directrices of the ellipse us \(\frac{x^2}{4}+\frac{y^2}{9}=1\) is
1 \(\frac{9}{\sqrt{5}}\)
2 \(\frac{18}{\sqrt{5}}\)
3 \(\frac{24}{\sqrt{5}}\)
4 None of these
Explanation:
B Given, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Here, \(\quad a^2=9\)
\(\mathrm{b}^2=4\)
Eccentricity, \(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{4}{9}}\)
\(\mathrm{e}=\frac{\sqrt{5}}{3}\)
Distance between directrix is \(=\frac{2 \mathrm{a}}{\mathrm{e}}\)
\(=\frac{2 \times 3}{\frac{\sqrt{5}}{3}}\)
\(=\frac{18}{\sqrt{5}}\)
