QBManager

Ellipse

120623 The line $a x + b y = 1$ cuts ellipse $c x^{2} + d y^{2} = 1$ only once if

1

{ca}^{2} + {db}^{2} = 1

2

\frac{c}{a^{2}} + \frac{d}{b^{2}} = 1

3

\frac{a^{2}}{c} + \frac{b^{2}}{d} = 1

4

{ac}^{2} + {bd}^{2} = 1

Explanation:

C Given, the line $a x + b y = 1$
i.e $y = - \frac{a}{b} x + \frac{1}{b}$ is tangent to
${cx}^{2} + {dy}^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{c}} + \frac{y^{2}}{\frac{1}{d}} = 1$
$∴ {(\frac{1}{b})}^{2} = (\frac{1}{c}) {(- \frac{a}{b})}^{2} + (\frac{1}{d})$
$1 = \frac{a^{2}}{c} + \frac{b^{2}}{d}$

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 $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$S$ is $(- ae, 0), T$ is $(ae, 0)$ and $B$ is $(0, b)$
original image
$SB = \sqrt{(0 + ae)^{2} + b^{2}}$
${SB}^{2} = {ST}^{2} \Rightarrow 4 a^{2} e^{2} = a^{2} e^{2} + b^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$3 a^{2} e^{2} = a^{2} (1 - e^{2}) = a^{2} - a^{2} e^{2}$
$4 a^{2} e^{2} = a^{2} \Rightarrow e^{2} = \frac{1}{4} \Rightarrow e = \frac{1}{2}$

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 m

2

8 / 3 m

3

7 / 2 m

4

2 m

Explanation:

B Given that, $2 a = 9 \Rightarrow a = \frac{9}{2}$ and $b = 3$
original image
The equation of the ellipse is
$\frac{x^{2}}{{(\frac{9}{2})}^{2}} + \frac{y^{2}}{9} = 1 .$
Where, centre is assumed as origin and base as $x$ -axis. Put $x = 2$ , we get :-
$\frac{16}{81} + \frac{y^{2}}{9} = 1 \Rightarrow y = \frac{\sqrt{65}}{3} = \frac{8}{3} m$ (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 ${SS}^{, as}$ diameter intersects the ellipse in real and distinct point, then the eccentricity $e$ of the ellipse satisfies

1

e = 1 / \sqrt{2}

2

e \in (1 / \sqrt{2}, 1)

3

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.
$\frac{x^{2}}{a^{2}} + \frac{a^{2} e^{2} - x^{2}}{b^{2}} = 1$
$x^{2} (\frac{1}{a^{2}} - \frac{1}{b^{2}}) = 1 - \frac{a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b^{2} - a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2}}) = b^{2} - a^{2} e^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$x = \pm \frac{a}{e} \sqrt{2 e^{2} - 1}$
This will give distinct value of $x$ , if
$2 e^{2} - 1 > 0 ie. e > \frac{1}{\sqrt{2}} .$
Hence, $e \in (1 / \sqrt{2}, 1)$

BITSAT-2016

Ellipse

120623 The line $a x + b y = 1$ cuts ellipse $c x^{2} + d y^{2} = 1$ only once if

1

{ca}^{2} + {db}^{2} = 1

2

\frac{c}{a^{2}} + \frac{d}{b^{2}} = 1

3

\frac{a^{2}}{c} + \frac{b^{2}}{d} = 1

4

{ac}^{2} + {bd}^{2} = 1

Explanation:

C Given, the line $a x + b y = 1$
i.e $y = - \frac{a}{b} x + \frac{1}{b}$ is tangent to
${cx}^{2} + {dy}^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{c}} + \frac{y^{2}}{\frac{1}{d}} = 1$
$∴ {(\frac{1}{b})}^{2} = (\frac{1}{c}) {(- \frac{a}{b})}^{2} + (\frac{1}{d})$
$1 = \frac{a^{2}}{c} + \frac{b^{2}}{d}$

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 $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$S$ is $(- ae, 0), T$ is $(ae, 0)$ and $B$ is $(0, b)$
original image
$SB = \sqrt{(0 + ae)^{2} + b^{2}}$
${SB}^{2} = {ST}^{2} \Rightarrow 4 a^{2} e^{2} = a^{2} e^{2} + b^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$3 a^{2} e^{2} = a^{2} (1 - e^{2}) = a^{2} - a^{2} e^{2}$
$4 a^{2} e^{2} = a^{2} \Rightarrow e^{2} = \frac{1}{4} \Rightarrow e = \frac{1}{2}$

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 m

2

8 / 3 m

3

7 / 2 m

4

2 m

Explanation:

B Given that, $2 a = 9 \Rightarrow a = \frac{9}{2}$ and $b = 3$
original image
The equation of the ellipse is
$\frac{x^{2}}{{(\frac{9}{2})}^{2}} + \frac{y^{2}}{9} = 1 .$
Where, centre is assumed as origin and base as $x$ -axis. Put $x = 2$ , we get :-
$\frac{16}{81} + \frac{y^{2}}{9} = 1 \Rightarrow y = \frac{\sqrt{65}}{3} = \frac{8}{3} m$ (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 ${SS}^{, as}$ diameter intersects the ellipse in real and distinct point, then the eccentricity $e$ of the ellipse satisfies

1

e = 1 / \sqrt{2}

2

e \in (1 / \sqrt{2}, 1)

3

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.
$\frac{x^{2}}{a^{2}} + \frac{a^{2} e^{2} - x^{2}}{b^{2}} = 1$
$x^{2} (\frac{1}{a^{2}} - \frac{1}{b^{2}}) = 1 - \frac{a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b^{2} - a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2}}) = b^{2} - a^{2} e^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$x = \pm \frac{a}{e} \sqrt{2 e^{2} - 1}$
This will give distinct value of $x$ , if
$2 e^{2} - 1 > 0 ie. e > \frac{1}{\sqrt{2}} .$
Hence, $e \in (1 / \sqrt{2}, 1)$

BITSAT-2016

Ellipse

120623 The line $a x + b y = 1$ cuts ellipse $c x^{2} + d y^{2} = 1$ only once if

1

{ca}^{2} + {db}^{2} = 1

2

\frac{c}{a^{2}} + \frac{d}{b^{2}} = 1

3

\frac{a^{2}}{c} + \frac{b^{2}}{d} = 1

4

{ac}^{2} + {bd}^{2} = 1

Explanation:

C Given, the line $a x + b y = 1$
i.e $y = - \frac{a}{b} x + \frac{1}{b}$ is tangent to
${cx}^{2} + {dy}^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{c}} + \frac{y^{2}}{\frac{1}{d}} = 1$
$∴ {(\frac{1}{b})}^{2} = (\frac{1}{c}) {(- \frac{a}{b})}^{2} + (\frac{1}{d})$
$1 = \frac{a^{2}}{c} + \frac{b^{2}}{d}$

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 $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$S$ is $(- ae, 0), T$ is $(ae, 0)$ and $B$ is $(0, b)$
original image
$SB = \sqrt{(0 + ae)^{2} + b^{2}}$
${SB}^{2} = {ST}^{2} \Rightarrow 4 a^{2} e^{2} = a^{2} e^{2} + b^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$3 a^{2} e^{2} = a^{2} (1 - e^{2}) = a^{2} - a^{2} e^{2}$
$4 a^{2} e^{2} = a^{2} \Rightarrow e^{2} = \frac{1}{4} \Rightarrow e = \frac{1}{2}$

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 m

2

8 / 3 m

3

7 / 2 m

4

2 m

Explanation:

B Given that, $2 a = 9 \Rightarrow a = \frac{9}{2}$ and $b = 3$
original image
The equation of the ellipse is
$\frac{x^{2}}{{(\frac{9}{2})}^{2}} + \frac{y^{2}}{9} = 1 .$
Where, centre is assumed as origin and base as $x$ -axis. Put $x = 2$ , we get :-
$\frac{16}{81} + \frac{y^{2}}{9} = 1 \Rightarrow y = \frac{\sqrt{65}}{3} = \frac{8}{3} m$ (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 ${SS}^{, as}$ diameter intersects the ellipse in real and distinct point, then the eccentricity $e$ of the ellipse satisfies

1

e = 1 / \sqrt{2}

2

e \in (1 / \sqrt{2}, 1)

3

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.
$\frac{x^{2}}{a^{2}} + \frac{a^{2} e^{2} - x^{2}}{b^{2}} = 1$
$x^{2} (\frac{1}{a^{2}} - \frac{1}{b^{2}}) = 1 - \frac{a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b^{2} - a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2}}) = b^{2} - a^{2} e^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$x = \pm \frac{a}{e} \sqrt{2 e^{2} - 1}$
This will give distinct value of $x$ , if
$2 e^{2} - 1 > 0 ie. e > \frac{1}{\sqrt{2}} .$
Hence, $e \in (1 / \sqrt{2}, 1)$

BITSAT-2016

Ellipse

120623 The line $a x + b y = 1$ cuts ellipse $c x^{2} + d y^{2} = 1$ only once if

1

{ca}^{2} + {db}^{2} = 1

2

\frac{c}{a^{2}} + \frac{d}{b^{2}} = 1

3

\frac{a^{2}}{c} + \frac{b^{2}}{d} = 1

4

{ac}^{2} + {bd}^{2} = 1

Explanation:

C Given, the line $a x + b y = 1$
i.e $y = - \frac{a}{b} x + \frac{1}{b}$ is tangent to
${cx}^{2} + {dy}^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{c}} + \frac{y^{2}}{\frac{1}{d}} = 1$
$∴ {(\frac{1}{b})}^{2} = (\frac{1}{c}) {(- \frac{a}{b})}^{2} + (\frac{1}{d})$
$1 = \frac{a^{2}}{c} + \frac{b^{2}}{d}$

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 $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$S$ is $(- ae, 0), T$ is $(ae, 0)$ and $B$ is $(0, b)$
original image
$SB = \sqrt{(0 + ae)^{2} + b^{2}}$
${SB}^{2} = {ST}^{2} \Rightarrow 4 a^{2} e^{2} = a^{2} e^{2} + b^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$3 a^{2} e^{2} = a^{2} (1 - e^{2}) = a^{2} - a^{2} e^{2}$
$4 a^{2} e^{2} = a^{2} \Rightarrow e^{2} = \frac{1}{4} \Rightarrow e = \frac{1}{2}$

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 m

2

8 / 3 m

3

7 / 2 m

4

2 m

Explanation:

B Given that, $2 a = 9 \Rightarrow a = \frac{9}{2}$ and $b = 3$
original image
The equation of the ellipse is
$\frac{x^{2}}{{(\frac{9}{2})}^{2}} + \frac{y^{2}}{9} = 1 .$
Where, centre is assumed as origin and base as $x$ -axis. Put $x = 2$ , we get :-
$\frac{16}{81} + \frac{y^{2}}{9} = 1 \Rightarrow y = \frac{\sqrt{65}}{3} = \frac{8}{3} m$ (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 ${SS}^{, as}$ diameter intersects the ellipse in real and distinct point, then the eccentricity $e$ of the ellipse satisfies

1

e = 1 / \sqrt{2}

2

e \in (1 / \sqrt{2}, 1)

3

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.
$\frac{x^{2}}{a^{2}} + \frac{a^{2} e^{2} - x^{2}}{b^{2}} = 1$
$x^{2} (\frac{1}{a^{2}} - \frac{1}{b^{2}}) = 1 - \frac{a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2} b^{2}}) = \frac{b^{2} - a^{2} e^{2}}{b^{2}}$
$x^{2} (\frac{b^{2} - a^{2}}{a^{2}}) = b^{2} - a^{2} e^{2}, {∵ b^{2} = a^{2} (1 - e)^{2}}$
$x = \pm \frac{a}{e} \sqrt{2 e^{2} - 1}$
This will give distinct value of $x$ , if
$2 e^{2} - 1 > 0 ie. e > \frac{1}{\sqrt{2}} .$
Hence, $e \in (1 / \sqrt{2}, 1)$

BITSAT-2016

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