C Given, the line \(a x+b y=1\)
i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to
\(\mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{\mathrm{c}}}+\frac{\mathrm{y}^2}{\frac{1}{\mathrm{~d}}}=1\)
\(\therefore\left(\frac{1}{\mathrm{~b}}\right)^2=\left(\frac{1}{\mathrm{c}}\right)\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)^2+\left(\frac{1}{\mathrm{~d}}\right)\)
\(1=\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}\)
BITSAT-2010
Ellipse
120624
\(S\) and \(T\) are the foci of an ellipse and \(B\) is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Let equation of ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{S}\) is \((-\mathrm{ae}, 0), \mathrm{T}\) is \((\mathrm{ae}, 0)\) and \(\mathrm{B}\) is \((0, \mathrm{~b})\)
\(\mathrm{SB}=\sqrt{(0+\mathrm{ae})^2+\mathrm{b}^2}\)
\(\mathrm{SB}^2=\mathrm{ST}^2 \Rightarrow 4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2, \quad\left\{\because \mathrm{b}^2=\mathrm{a}^2(1-\mathrm{e})^2\right\}\)
\(3 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{a}^2-\mathrm{a}^2 \mathrm{e}^2\)
\(4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \Rightarrow \mathrm{e}^2=\frac{1}{4} \Rightarrow \mathrm{e}=\frac{1}{2}\)
BITSAT-2013
Ellipse
120625
An arch of a bridge is semi-elliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is
1 \(11 / 4 \mathrm{~m}\)
2 \(8 / 3 \mathrm{~m}\)
3 \(7 / 2 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
B Given that, \(2 \mathrm{a}=9 \Rightarrow \mathrm{a}=\frac{9}{2}\) and \(\mathrm{b}=3\)
The equation of the ellipse is
\(\frac{\mathrm{x}^2}{\left(\frac{9}{2}\right)^2}+\frac{\mathrm{y}^2}{9}=1 \text {. }\)
Where, centre is assumed as origin and base as \(\mathrm{x}\)-axis. Put \(x=2\), we get :-
\(\frac{16}{81}+\frac{y^2}{9}=1 \Rightarrow \mathrm{y}=\frac{\sqrt{65}}{3}=\frac{8}{3} \mathrm{~m}\)(approximately)
BITSAT-2014
Ellipse
120626
Let \(S\) and \(S '\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on \(\mathrm{SS}^{\text {, as }}\) diameter intersects the ellipse in real and distinct point, then the eccentricity \(e\) of the ellipse satisfies
1 \(\mathrm{e}=1 / \sqrt{2}\)
2 \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
3 \(\mathrm{e} \in(0,1 \sqrt{2})\)
4 None of these
Explanation:
B The equation of the circle described on SS' as a diameter is
\((x-a e)(x+a e)+(y-0)(y-0)=0\)
\(x^2-a^2 e^2+y^2=0\)
\(x^2+y^2=a^2 e^2\)
The abscissae of the points of intersection of the ellipse and this circle are the roots of the equation.
\(\frac{x^2}{a^2}+\frac{a^2 e^2-x^2}{b^2}=1\)
\(x^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=1-\frac{a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2}\right)=b^2-a^2 e^2, \quad\left\{\because b^2=a^2(1-e)^2\right\}\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
This will give distinct value of \(\mathrm{x}\), if
\(2 \mathrm{e}^2-1>0 \text { ie. e }>\frac{1}{\sqrt{2}} \text {. }\)
Hence, \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
C Given, the line \(a x+b y=1\)
i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to
\(\mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{\mathrm{c}}}+\frac{\mathrm{y}^2}{\frac{1}{\mathrm{~d}}}=1\)
\(\therefore\left(\frac{1}{\mathrm{~b}}\right)^2=\left(\frac{1}{\mathrm{c}}\right)\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)^2+\left(\frac{1}{\mathrm{~d}}\right)\)
\(1=\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}\)
BITSAT-2010
Ellipse
120624
\(S\) and \(T\) are the foci of an ellipse and \(B\) is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Let equation of ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{S}\) is \((-\mathrm{ae}, 0), \mathrm{T}\) is \((\mathrm{ae}, 0)\) and \(\mathrm{B}\) is \((0, \mathrm{~b})\)
\(\mathrm{SB}=\sqrt{(0+\mathrm{ae})^2+\mathrm{b}^2}\)
\(\mathrm{SB}^2=\mathrm{ST}^2 \Rightarrow 4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2, \quad\left\{\because \mathrm{b}^2=\mathrm{a}^2(1-\mathrm{e})^2\right\}\)
\(3 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{a}^2-\mathrm{a}^2 \mathrm{e}^2\)
\(4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \Rightarrow \mathrm{e}^2=\frac{1}{4} \Rightarrow \mathrm{e}=\frac{1}{2}\)
BITSAT-2013
Ellipse
120625
An arch of a bridge is semi-elliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is
1 \(11 / 4 \mathrm{~m}\)
2 \(8 / 3 \mathrm{~m}\)
3 \(7 / 2 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
B Given that, \(2 \mathrm{a}=9 \Rightarrow \mathrm{a}=\frac{9}{2}\) and \(\mathrm{b}=3\)
The equation of the ellipse is
\(\frac{\mathrm{x}^2}{\left(\frac{9}{2}\right)^2}+\frac{\mathrm{y}^2}{9}=1 \text {. }\)
Where, centre is assumed as origin and base as \(\mathrm{x}\)-axis. Put \(x=2\), we get :-
\(\frac{16}{81}+\frac{y^2}{9}=1 \Rightarrow \mathrm{y}=\frac{\sqrt{65}}{3}=\frac{8}{3} \mathrm{~m}\)(approximately)
BITSAT-2014
Ellipse
120626
Let \(S\) and \(S '\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on \(\mathrm{SS}^{\text {, as }}\) diameter intersects the ellipse in real and distinct point, then the eccentricity \(e\) of the ellipse satisfies
1 \(\mathrm{e}=1 / \sqrt{2}\)
2 \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
3 \(\mathrm{e} \in(0,1 \sqrt{2})\)
4 None of these
Explanation:
B The equation of the circle described on SS' as a diameter is
\((x-a e)(x+a e)+(y-0)(y-0)=0\)
\(x^2-a^2 e^2+y^2=0\)
\(x^2+y^2=a^2 e^2\)
The abscissae of the points of intersection of the ellipse and this circle are the roots of the equation.
\(\frac{x^2}{a^2}+\frac{a^2 e^2-x^2}{b^2}=1\)
\(x^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=1-\frac{a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2}\right)=b^2-a^2 e^2, \quad\left\{\because b^2=a^2(1-e)^2\right\}\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
This will give distinct value of \(\mathrm{x}\), if
\(2 \mathrm{e}^2-1>0 \text { ie. e }>\frac{1}{\sqrt{2}} \text {. }\)
Hence, \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
C Given, the line \(a x+b y=1\)
i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to
\(\mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{\mathrm{c}}}+\frac{\mathrm{y}^2}{\frac{1}{\mathrm{~d}}}=1\)
\(\therefore\left(\frac{1}{\mathrm{~b}}\right)^2=\left(\frac{1}{\mathrm{c}}\right)\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)^2+\left(\frac{1}{\mathrm{~d}}\right)\)
\(1=\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}\)
BITSAT-2010
Ellipse
120624
\(S\) and \(T\) are the foci of an ellipse and \(B\) is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Let equation of ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{S}\) is \((-\mathrm{ae}, 0), \mathrm{T}\) is \((\mathrm{ae}, 0)\) and \(\mathrm{B}\) is \((0, \mathrm{~b})\)
\(\mathrm{SB}=\sqrt{(0+\mathrm{ae})^2+\mathrm{b}^2}\)
\(\mathrm{SB}^2=\mathrm{ST}^2 \Rightarrow 4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2, \quad\left\{\because \mathrm{b}^2=\mathrm{a}^2(1-\mathrm{e})^2\right\}\)
\(3 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{a}^2-\mathrm{a}^2 \mathrm{e}^2\)
\(4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \Rightarrow \mathrm{e}^2=\frac{1}{4} \Rightarrow \mathrm{e}=\frac{1}{2}\)
BITSAT-2013
Ellipse
120625
An arch of a bridge is semi-elliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is
1 \(11 / 4 \mathrm{~m}\)
2 \(8 / 3 \mathrm{~m}\)
3 \(7 / 2 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
B Given that, \(2 \mathrm{a}=9 \Rightarrow \mathrm{a}=\frac{9}{2}\) and \(\mathrm{b}=3\)
The equation of the ellipse is
\(\frac{\mathrm{x}^2}{\left(\frac{9}{2}\right)^2}+\frac{\mathrm{y}^2}{9}=1 \text {. }\)
Where, centre is assumed as origin and base as \(\mathrm{x}\)-axis. Put \(x=2\), we get :-
\(\frac{16}{81}+\frac{y^2}{9}=1 \Rightarrow \mathrm{y}=\frac{\sqrt{65}}{3}=\frac{8}{3} \mathrm{~m}\)(approximately)
BITSAT-2014
Ellipse
120626
Let \(S\) and \(S '\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on \(\mathrm{SS}^{\text {, as }}\) diameter intersects the ellipse in real and distinct point, then the eccentricity \(e\) of the ellipse satisfies
1 \(\mathrm{e}=1 / \sqrt{2}\)
2 \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
3 \(\mathrm{e} \in(0,1 \sqrt{2})\)
4 None of these
Explanation:
B The equation of the circle described on SS' as a diameter is
\((x-a e)(x+a e)+(y-0)(y-0)=0\)
\(x^2-a^2 e^2+y^2=0\)
\(x^2+y^2=a^2 e^2\)
The abscissae of the points of intersection of the ellipse and this circle are the roots of the equation.
\(\frac{x^2}{a^2}+\frac{a^2 e^2-x^2}{b^2}=1\)
\(x^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=1-\frac{a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2}\right)=b^2-a^2 e^2, \quad\left\{\because b^2=a^2(1-e)^2\right\}\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
This will give distinct value of \(\mathrm{x}\), if
\(2 \mathrm{e}^2-1>0 \text { ie. e }>\frac{1}{\sqrt{2}} \text {. }\)
Hence, \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
C Given, the line \(a x+b y=1\)
i.e \(y=-\frac{a}{b} x+\frac{1}{b}\) is tangent to
\(\mathrm{cx}^2+\mathrm{dy}^2=1 \Rightarrow \frac{\mathrm{x}^2}{\frac{1}{\mathrm{c}}}+\frac{\mathrm{y}^2}{\frac{1}{\mathrm{~d}}}=1\)
\(\therefore\left(\frac{1}{\mathrm{~b}}\right)^2=\left(\frac{1}{\mathrm{c}}\right)\left(-\frac{\mathrm{a}}{\mathrm{b}}\right)^2+\left(\frac{1}{\mathrm{~d}}\right)\)
\(1=\frac{\mathrm{a}^2}{\mathrm{c}}+\frac{\mathrm{b}^2}{\mathrm{~d}}\)
BITSAT-2010
Ellipse
120624
\(S\) and \(T\) are the foci of an ellipse and \(B\) is an end of the minor axis. If STB is an equilateral triangle, then the eccentricity of the ellipse is
1 \(\frac{1}{4}\)
2 \(\frac{1}{3}\)
3 \(\frac{1}{2}\)
4 \(\frac{2}{3}\)
Explanation:
C Let equation of ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
\(\mathrm{S}\) is \((-\mathrm{ae}, 0), \mathrm{T}\) is \((\mathrm{ae}, 0)\) and \(\mathrm{B}\) is \((0, \mathrm{~b})\)
\(\mathrm{SB}=\sqrt{(0+\mathrm{ae})^2+\mathrm{b}^2}\)
\(\mathrm{SB}^2=\mathrm{ST}^2 \Rightarrow 4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \mathrm{e}^2+\mathrm{b}^2, \quad\left\{\because \mathrm{b}^2=\mathrm{a}^2(1-\mathrm{e})^2\right\}\)
\(3 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{a}^2-\mathrm{a}^2 \mathrm{e}^2\)
\(4 \mathrm{a}^2 \mathrm{e}^2=\mathrm{a}^2 \Rightarrow \mathrm{e}^2=\frac{1}{4} \Rightarrow \mathrm{e}=\frac{1}{2}\)
BITSAT-2013
Ellipse
120625
An arch of a bridge is semi-elliptical with major axis horizontal. If the length the base is 9 meter and the highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch. 2 meter from the centre of the base is
1 \(11 / 4 \mathrm{~m}\)
2 \(8 / 3 \mathrm{~m}\)
3 \(7 / 2 \mathrm{~m}\)
4 \(2 \mathrm{~m}\)
Explanation:
B Given that, \(2 \mathrm{a}=9 \Rightarrow \mathrm{a}=\frac{9}{2}\) and \(\mathrm{b}=3\)
The equation of the ellipse is
\(\frac{\mathrm{x}^2}{\left(\frac{9}{2}\right)^2}+\frac{\mathrm{y}^2}{9}=1 \text {. }\)
Where, centre is assumed as origin and base as \(\mathrm{x}\)-axis. Put \(x=2\), we get :-
\(\frac{16}{81}+\frac{y^2}{9}=1 \Rightarrow \mathrm{y}=\frac{\sqrt{65}}{3}=\frac{8}{3} \mathrm{~m}\)(approximately)
BITSAT-2014
Ellipse
120626
Let \(S\) and \(S '\) be two foci of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If a circle described on \(\mathrm{SS}^{\text {, as }}\) diameter intersects the ellipse in real and distinct point, then the eccentricity \(e\) of the ellipse satisfies
1 \(\mathrm{e}=1 / \sqrt{2}\)
2 \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
3 \(\mathrm{e} \in(0,1 \sqrt{2})\)
4 None of these
Explanation:
B The equation of the circle described on SS' as a diameter is
\((x-a e)(x+a e)+(y-0)(y-0)=0\)
\(x^2-a^2 e^2+y^2=0\)
\(x^2+y^2=a^2 e^2\)
The abscissae of the points of intersection of the ellipse and this circle are the roots of the equation.
\(\frac{x^2}{a^2}+\frac{a^2 e^2-x^2}{b^2}=1\)
\(x^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=1-\frac{a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2 b^2}\right)=\frac{b^2-a^2 e^2}{b^2}\)
\(x^2\left(\frac{b^2-a^2}{a^2}\right)=b^2-a^2 e^2, \quad\left\{\because b^2=a^2(1-e)^2\right\}\)
\(x= \pm \frac{a}{e} \sqrt{2 e^2-1}\)
This will give distinct value of \(\mathrm{x}\), if
\(2 \mathrm{e}^2-1>0 \text { ie. e }>\frac{1}{\sqrt{2}} \text {. }\)
Hence, \(\mathrm{e} \in(1 / \sqrt{2}, 1)\)
