QBManager

Ellipse

120665 The normal drawn at the point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$ to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$ intersects its major axis at the point

1

(0, \sqrt{\frac{2}{7}})

2

(- \sqrt{\frac{2}{9}}, 0)

3

(0, - \sqrt{\frac{2}{7}})

4

(\sqrt{\frac{2}{9}}, 0)

Explanation:

D Given, the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$
Differentiating w.r.t. $x$ , we get -
$\frac{2 x}{9} + \frac{2 y}{7} \frac{d y}{d x} = 0$
$∴ \frac{d y}{d x} = \frac{- 2 x / 9}{2 y / 7} = \frac{- x}{y} (\frac{7}{9})$
Now, $\frac{dy}{dx}$ at point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$= \frac{- \sqrt{9} \cos \frac{π}{4}}{\sqrt{7} \sin \frac{π}{4}} \cdot \frac{7}{9} = - \sqrt{\frac{7}{9}}$
$∴$ Slope of the normal, $m = \frac{3}{\sqrt{7}}$
$∴$ Equation of this normal $y = \frac{3}{\sqrt{7}} x + c$
But this line passes through $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$∴ \sqrt{7} \cdot \sin \frac{π}{4} = \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{π}{4} + c$
$∴ c = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}$
$= \frac{1}{\sqrt{2}} [\frac{7 - 9}{\sqrt{7}}] = \frac{- 2}{\sqrt{14}}$
$∴$ Equation of the normal is $y = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
Since this line intersects its major axis
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴$ The point of intersection is $(\sqrt{\frac{2}{9}}, 0)$ .

TS EAMCET-04.05.2019

Ellipse

120666 The locus of the mid-points of the portion of the tangents of the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ intercepted between the coordinate axes is

1

\frac{1}{4 x^{2}} + \frac{1}{2 y^{2}} = 1

2

2 x^{2} + y^{2} = 4

3

\frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1

4

x^{2} + 2 y^{2} = 4

Explanation:

C Given, equation of ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$
Here, $a^{2} = 2, b^{2} = 1$
Equaiton of tangent of ellipse are
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
The $x$ -intercept and y-intercept are $(\sqrt{2} \sec θ, 0)$ and $(0$ , $cosec θ$ ) respectively.
Let, (h, k) mid-point of intercept -
$h = \frac{\sqrt{2} \sec θ}{2}, k = \frac{cosec θ}{2}$
$\cos θ = \frac{1}{\sqrt{2} h} and \sin θ = \frac{1}{2 k}$
Squaring and adding, we get -
$\frac{1}{2 h^{2}} + \frac{1}{4 k^{2}} = 1$
$∴ locus be \frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1$

TS EAMCET-03.05.2019

Ellipse

120667 If a circle $(x - 1)^{2} + y^{2} = r^{2}$ touches the ellipse $x^{2}$ $+ 4 y^{2} = 16$ internally, then $r =$

1

\sqrt{\frac{11}{3}}

2

\frac{11}{3}

3

\sqrt{\frac{15}{2}}

4 2

Explanation:

A Given, the circle $(x - 1)^{2} + y^{2} = r^{2}$
Centre of the circle is $(1, 0)$
Given the equation of the ellipse $= x^{2} + 4 y^{2} = 16$
Point of the ellipse $= (4 \cos θ, 2 \sin θ)$
Equation of normal of ellipse
$\Rightarrow$ ax $\sec θ -$ by $cosec θ = a^{2} - b^{2}$
$\Rightarrow 4 x \sec θ - 2 y cosec θ = 12$
It passes through $(1, 0)$
$\Rightarrow \sec θ = 3$
$\Rightarrow \sin θ = \frac{2 \sqrt{2}}{3} \cos θ = \frac{1}{3}$
Now, point will be $(\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$Radius = Distance of (1, 0) and (\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$= \sqrt{{(1 - \frac{4}{3})}^{2} + {(0 - \frac{4 \sqrt{2}}{3})}^{2}}$
$= \sqrt{\frac{1}{9} + \frac{32}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}}$

TS EAMCET-05.08.2021

Ellipse

120668 The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse $S = \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is

1 96

2 16

3 128

4 64

Explanation:

D Given ellipse, $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$
original image
End point of latus rectum $= (\pm ae, \frac{2 b^{2}}{a}) = (2, 3)$
Equation of tangent at $(2, 3)$ of ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is,
$\frac{2 x}{16} + \frac{3 y}{12} = 1 \Rightarrow \frac{x}{8} + \frac{y}{4} = 1$
Area of quadrilateral $ABCD = 4 \times \frac{1}{2} \times OA \times OB$
$= 4 \times \frac{1}{2} \times 8 \times 4 = 64$

TS EAMCET 14.09.2020

Ellipse

120665 The normal drawn at the point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$ to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$ intersects its major axis at the point

1

(0, \sqrt{\frac{2}{7}})

2

(- \sqrt{\frac{2}{9}}, 0)

3

(0, - \sqrt{\frac{2}{7}})

4

(\sqrt{\frac{2}{9}}, 0)

Explanation:

D Given, the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$
Differentiating w.r.t. $x$ , we get -
$\frac{2 x}{9} + \frac{2 y}{7} \frac{d y}{d x} = 0$
$∴ \frac{d y}{d x} = \frac{- 2 x / 9}{2 y / 7} = \frac{- x}{y} (\frac{7}{9})$
Now, $\frac{dy}{dx}$ at point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$= \frac{- \sqrt{9} \cos \frac{π}{4}}{\sqrt{7} \sin \frac{π}{4}} \cdot \frac{7}{9} = - \sqrt{\frac{7}{9}}$
$∴$ Slope of the normal, $m = \frac{3}{\sqrt{7}}$
$∴$ Equation of this normal $y = \frac{3}{\sqrt{7}} x + c$
But this line passes through $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$∴ \sqrt{7} \cdot \sin \frac{π}{4} = \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{π}{4} + c$
$∴ c = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}$
$= \frac{1}{\sqrt{2}} [\frac{7 - 9}{\sqrt{7}}] = \frac{- 2}{\sqrt{14}}$
$∴$ Equation of the normal is $y = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
Since this line intersects its major axis
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴$ The point of intersection is $(\sqrt{\frac{2}{9}}, 0)$ .

TS EAMCET-04.05.2019

Ellipse

120666 The locus of the mid-points of the portion of the tangents of the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ intercepted between the coordinate axes is

1

\frac{1}{4 x^{2}} + \frac{1}{2 y^{2}} = 1

2

2 x^{2} + y^{2} = 4

3

\frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1

4

x^{2} + 2 y^{2} = 4

Explanation:

C Given, equation of ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$
Here, $a^{2} = 2, b^{2} = 1$
Equaiton of tangent of ellipse are
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
The $x$ -intercept and y-intercept are $(\sqrt{2} \sec θ, 0)$ and $(0$ , $cosec θ$ ) respectively.
Let, (h, k) mid-point of intercept -
$h = \frac{\sqrt{2} \sec θ}{2}, k = \frac{cosec θ}{2}$
$\cos θ = \frac{1}{\sqrt{2} h} and \sin θ = \frac{1}{2 k}$
Squaring and adding, we get -
$\frac{1}{2 h^{2}} + \frac{1}{4 k^{2}} = 1$
$∴ locus be \frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1$

TS EAMCET-03.05.2019

Ellipse

120667 If a circle $(x - 1)^{2} + y^{2} = r^{2}$ touches the ellipse $x^{2}$ $+ 4 y^{2} = 16$ internally, then $r =$

1

\sqrt{\frac{11}{3}}

2

\frac{11}{3}

3

\sqrt{\frac{15}{2}}

4 2

Explanation:

A Given, the circle $(x - 1)^{2} + y^{2} = r^{2}$
Centre of the circle is $(1, 0)$
Given the equation of the ellipse $= x^{2} + 4 y^{2} = 16$
Point of the ellipse $= (4 \cos θ, 2 \sin θ)$
Equation of normal of ellipse
$\Rightarrow$ ax $\sec θ -$ by $cosec θ = a^{2} - b^{2}$
$\Rightarrow 4 x \sec θ - 2 y cosec θ = 12$
It passes through $(1, 0)$
$\Rightarrow \sec θ = 3$
$\Rightarrow \sin θ = \frac{2 \sqrt{2}}{3} \cos θ = \frac{1}{3}$
Now, point will be $(\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$Radius = Distance of (1, 0) and (\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$= \sqrt{{(1 - \frac{4}{3})}^{2} + {(0 - \frac{4 \sqrt{2}}{3})}^{2}}$
$= \sqrt{\frac{1}{9} + \frac{32}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}}$

TS EAMCET-05.08.2021

Ellipse

120668 The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse $S = \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is

1 96

2 16

3 128

4 64

Explanation:

D Given ellipse, $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$
original image
End point of latus rectum $= (\pm ae, \frac{2 b^{2}}{a}) = (2, 3)$
Equation of tangent at $(2, 3)$ of ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is,
$\frac{2 x}{16} + \frac{3 y}{12} = 1 \Rightarrow \frac{x}{8} + \frac{y}{4} = 1$
Area of quadrilateral $ABCD = 4 \times \frac{1}{2} \times OA \times OB$
$= 4 \times \frac{1}{2} \times 8 \times 4 = 64$

TS EAMCET 14.09.2020

Ellipse

120665 The normal drawn at the point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$ to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$ intersects its major axis at the point

1

(0, \sqrt{\frac{2}{7}})

2

(- \sqrt{\frac{2}{9}}, 0)

3

(0, - \sqrt{\frac{2}{7}})

4

(\sqrt{\frac{2}{9}}, 0)

Explanation:

D Given, the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$
Differentiating w.r.t. $x$ , we get -
$\frac{2 x}{9} + \frac{2 y}{7} \frac{d y}{d x} = 0$
$∴ \frac{d y}{d x} = \frac{- 2 x / 9}{2 y / 7} = \frac{- x}{y} (\frac{7}{9})$
Now, $\frac{dy}{dx}$ at point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$= \frac{- \sqrt{9} \cos \frac{π}{4}}{\sqrt{7} \sin \frac{π}{4}} \cdot \frac{7}{9} = - \sqrt{\frac{7}{9}}$
$∴$ Slope of the normal, $m = \frac{3}{\sqrt{7}}$
$∴$ Equation of this normal $y = \frac{3}{\sqrt{7}} x + c$
But this line passes through $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$∴ \sqrt{7} \cdot \sin \frac{π}{4} = \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{π}{4} + c$
$∴ c = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}$
$= \frac{1}{\sqrt{2}} [\frac{7 - 9}{\sqrt{7}}] = \frac{- 2}{\sqrt{14}}$
$∴$ Equation of the normal is $y = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
Since this line intersects its major axis
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴$ The point of intersection is $(\sqrt{\frac{2}{9}}, 0)$ .

TS EAMCET-04.05.2019

Ellipse

120666 The locus of the mid-points of the portion of the tangents of the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ intercepted between the coordinate axes is

1

\frac{1}{4 x^{2}} + \frac{1}{2 y^{2}} = 1

2

2 x^{2} + y^{2} = 4

3

\frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1

4

x^{2} + 2 y^{2} = 4

Explanation:

C Given, equation of ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$
Here, $a^{2} = 2, b^{2} = 1$
Equaiton of tangent of ellipse are
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
The $x$ -intercept and y-intercept are $(\sqrt{2} \sec θ, 0)$ and $(0$ , $cosec θ$ ) respectively.
Let, (h, k) mid-point of intercept -
$h = \frac{\sqrt{2} \sec θ}{2}, k = \frac{cosec θ}{2}$
$\cos θ = \frac{1}{\sqrt{2} h} and \sin θ = \frac{1}{2 k}$
Squaring and adding, we get -
$\frac{1}{2 h^{2}} + \frac{1}{4 k^{2}} = 1$
$∴ locus be \frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1$

TS EAMCET-03.05.2019

Ellipse

120667 If a circle $(x - 1)^{2} + y^{2} = r^{2}$ touches the ellipse $x^{2}$ $+ 4 y^{2} = 16$ internally, then $r =$

1

\sqrt{\frac{11}{3}}

2

\frac{11}{3}

3

\sqrt{\frac{15}{2}}

4 2

Explanation:

A Given, the circle $(x - 1)^{2} + y^{2} = r^{2}$
Centre of the circle is $(1, 0)$
Given the equation of the ellipse $= x^{2} + 4 y^{2} = 16$
Point of the ellipse $= (4 \cos θ, 2 \sin θ)$
Equation of normal of ellipse
$\Rightarrow$ ax $\sec θ -$ by $cosec θ = a^{2} - b^{2}$
$\Rightarrow 4 x \sec θ - 2 y cosec θ = 12$
It passes through $(1, 0)$
$\Rightarrow \sec θ = 3$
$\Rightarrow \sin θ = \frac{2 \sqrt{2}}{3} \cos θ = \frac{1}{3}$
Now, point will be $(\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$Radius = Distance of (1, 0) and (\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$= \sqrt{{(1 - \frac{4}{3})}^{2} + {(0 - \frac{4 \sqrt{2}}{3})}^{2}}$
$= \sqrt{\frac{1}{9} + \frac{32}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}}$

TS EAMCET-05.08.2021

Ellipse

120668 The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse $S = \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is

1 96

2 16

3 128

4 64

Explanation:

D Given ellipse, $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$
original image
End point of latus rectum $= (\pm ae, \frac{2 b^{2}}{a}) = (2, 3)$
Equation of tangent at $(2, 3)$ of ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is,
$\frac{2 x}{16} + \frac{3 y}{12} = 1 \Rightarrow \frac{x}{8} + \frac{y}{4} = 1$
Area of quadrilateral $ABCD = 4 \times \frac{1}{2} \times OA \times OB$
$= 4 \times \frac{1}{2} \times 8 \times 4 = 64$

TS EAMCET 14.09.2020

Ellipse

120665 The normal drawn at the point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$ to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$ intersects its major axis at the point

1

(0, \sqrt{\frac{2}{7}})

2

(- \sqrt{\frac{2}{9}}, 0)

3

(0, - \sqrt{\frac{2}{7}})

4

(\sqrt{\frac{2}{9}}, 0)

Explanation:

D Given, the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{7} = 1$
Differentiating w.r.t. $x$ , we get -
$\frac{2 x}{9} + \frac{2 y}{7} \frac{d y}{d x} = 0$
$∴ \frac{d y}{d x} = \frac{- 2 x / 9}{2 y / 7} = \frac{- x}{y} (\frac{7}{9})$
Now, $\frac{dy}{dx}$ at point $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$= \frac{- \sqrt{9} \cos \frac{π}{4}}{\sqrt{7} \sin \frac{π}{4}} \cdot \frac{7}{9} = - \sqrt{\frac{7}{9}}$
$∴$ Slope of the normal, $m = \frac{3}{\sqrt{7}}$
$∴$ Equation of this normal $y = \frac{3}{\sqrt{7}} x + c$
But this line passes through $(\sqrt{9} \cos \frac{π}{4}, \sqrt{7} \sin \frac{π}{4})$
$∴ \sqrt{7} \cdot \sin \frac{π}{4} = \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \cos \frac{π}{4} + c$
$∴ c = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3}{\sqrt{7}} \cdot \sqrt{9} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{7}}{\sqrt{2}} - \frac{3 \times 3}{\sqrt{2} \cdot \sqrt{7}}$
$= \frac{1}{\sqrt{2}} [\frac{7 - 9}{\sqrt{7}}] = \frac{- 2}{\sqrt{14}}$
$∴$ Equation of the normal is $y = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
Since this line intersects its major axis
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴ y = 0$
$∴ 0 = \frac{3}{\sqrt{7}} x - \frac{2}{\sqrt{14}}$
$x = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{3} = \frac{2 \times \sqrt{7}}{\sqrt{2} \times \sqrt{7} \times 3} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$
$∴$ The point of intersection is $(\sqrt{\frac{2}{9}}, 0)$ .

TS EAMCET-04.05.2019

Ellipse

120666 The locus of the mid-points of the portion of the tangents of the ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ intercepted between the coordinate axes is

1

\frac{1}{4 x^{2}} + \frac{1}{2 y^{2}} = 1

2

2 x^{2} + y^{2} = 4

3

\frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1

4

x^{2} + 2 y^{2} = 4

Explanation:

C Given, equation of ellipse $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$
Here, $a^{2} = 2, b^{2} = 1$
Equaiton of tangent of ellipse are
$\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$
The $x$ -intercept and y-intercept are $(\sqrt{2} \sec θ, 0)$ and $(0$ , $cosec θ$ ) respectively.
Let, (h, k) mid-point of intercept -
$h = \frac{\sqrt{2} \sec θ}{2}, k = \frac{cosec θ}{2}$
$\cos θ = \frac{1}{\sqrt{2} h} and \sin θ = \frac{1}{2 k}$
Squaring and adding, we get -
$\frac{1}{2 h^{2}} + \frac{1}{4 k^{2}} = 1$
$∴ locus be \frac{1}{2 x^{2}} + \frac{1}{4 y^{2}} = 1$

TS EAMCET-03.05.2019

Ellipse

120667 If a circle $(x - 1)^{2} + y^{2} = r^{2}$ touches the ellipse $x^{2}$ $+ 4 y^{2} = 16$ internally, then $r =$

1

\sqrt{\frac{11}{3}}

2

\frac{11}{3}

3

\sqrt{\frac{15}{2}}

4 2

Explanation:

A Given, the circle $(x - 1)^{2} + y^{2} = r^{2}$
Centre of the circle is $(1, 0)$
Given the equation of the ellipse $= x^{2} + 4 y^{2} = 16$
Point of the ellipse $= (4 \cos θ, 2 \sin θ)$
Equation of normal of ellipse
$\Rightarrow$ ax $\sec θ -$ by $cosec θ = a^{2} - b^{2}$
$\Rightarrow 4 x \sec θ - 2 y cosec θ = 12$
It passes through $(1, 0)$
$\Rightarrow \sec θ = 3$
$\Rightarrow \sin θ = \frac{2 \sqrt{2}}{3} \cos θ = \frac{1}{3}$
Now, point will be $(\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$Radius = Distance of (1, 0) and (\frac{4}{3}, \frac{4 \sqrt{2}}{3})$
$= \sqrt{{(1 - \frac{4}{3})}^{2} + {(0 - \frac{4 \sqrt{2}}{3})}^{2}}$
$= \sqrt{\frac{1}{9} + \frac{32}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}}$

TS EAMCET-05.08.2021

Ellipse

120668 The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse $S = \frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is

1 96

2 16

3 128

4 64

Explanation:

D Given ellipse, $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$
original image
End point of latus rectum $= (\pm ae, \frac{2 b^{2}}{a}) = (2, 3)$
Equation of tangent at $(2, 3)$ of ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{12} = 1$ is,
$\frac{2 x}{16} + \frac{3 y}{12} = 1 \Rightarrow \frac{x}{8} + \frac{y}{4} = 1$
Area of quadrilateral $ABCD = 4 \times \frac{1}{2} \times OA \times OB$
$= 4 \times \frac{1}{2} \times 8 \times 4 = 64$

TS EAMCET 14.09.2020

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