QBManager

Ellipse

120642 If the tangent at a point $P$ one the parabola $y^{2} =$ $3 x$ is parallel to the line $x + 2 y = 1$ and tangents at the point $Q$ and $R$ on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ are perpendicular to the line $x - y =$ 2, then the area of the triangle $P Q R$ is:

1

3 \sqrt{5}

2

\frac{9}{\sqrt{5}}

3

\frac{3}{2} \sqrt{5}

4

5 \sqrt{3}

Explanation:

A Given,
$y^{2} = 3 x$
Tangent $P (x_{1}, y_{1})$ is parallel to, $x + 2 y = 1$
Then slope at $P = \frac{- 1}{2}$
$2 y \frac{d y}{d x} = 3$
$\frac{d y}{d x} = \frac{3}{2 y} = \frac{- 1}{2}$
$y = - 3$
$∴ = (3, - 3)$
Similarly, $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $R = (\frac{- 4}{\sqrt{5}}, \frac{- 1}{\sqrt{5}})$
Area of $△ PQR = \frac{1}{2} | \begin{array}{ccc} 3 - 31 \\ \frac{4}{\sqrt{5}} \frac{1}{\sqrt{5}} 1 \\ \frac{- 4}{\sqrt{5}} \frac{- 1}{\sqrt{5}} 1 \end{array} |$
$= \frac{1}{2} [3 (\frac{2}{\sqrt{5}}) + 3 (\frac{8}{\sqrt{5}}) + 0]$
$= \frac{30}{2 \sqrt{5}}$
$= 3 \sqrt{5}$

JEE Main-29.01.2023

Ellipse

120643 Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , On comparing standard ellipse equation, $a^{2} = 19, b^{2} = 15$ , Equation of tangent of ellipse is -, $y = mx + \sqrt{19 m^{2} + 15}$ , $mx - y \pm \sqrt{19 m^{2} + 15} = 0$ , Perpendicular distance from centre of circle to tangent $=$ , 4, $| \frac{\pm \sqrt{19 m^{2} + 15}}{\sqrt{m^{2} + 1}} | = 4$ , $19 m^{2} + 15 = 16 m^{2} + 16$ , $m = \pm \frac{1}{\sqrt{3}}$ , $\tan θ = m = \frac{1}{\sqrt{3}}$ , $θ = \frac{π}{6}$ with $x$ -axis or $\frac{π}{3}$ with $y$ -axis, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{π}{3}$ ., 911. If the tangent drawn to the parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is the normal to the ellipse $4 x^{2} + 5 y^{2} = 20$ at $(\sqrt{5} \cos θ, 2 \sin θ)$ , then,

1

\frac{π}{4}

2

\frac{π}{3}

3

\frac{π}{12}

4

\frac{π}{6}

Explanation:

B Given,
$15 x^{2} + 19 y^{2} = 285$
$\frac{x^{2}}{19} + \frac{y^{2}}{15} = 1$
Ans: a
Exp: (A) : Given, tangent to parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is $y \cdot y_{1} = 2 (x + x_{1})$
$y \cdot 2 t = 2 (x + t^{2})$
$y t = x + t^{2}$
$y = \frac{x}{t} + t$
Normal to the ellipse, $4 x^{2} + 5 y^{2} = 20$
$\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1 at (\sqrt{5} \cos θ, 2 \sin θ)$
Slope of normal $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$ at $(\sqrt{5} \cos θ, 2 \sin θ)$
$∴$ Equation of line is -
$y - 2 \sin θ = \frac{\sqrt{5}}{2} \tan θ (x - \sqrt{5} \cos θ)$
$y = \frac{\sqrt{5}}{2} \tan θ \times x - \frac{\sin θ}{2}$
By comparing equation (i) and equation (ii), we get -
$t = - \frac{\sin θ}{2}$
$\sin θ = - 2 t$
$\frac{1}{t} = \frac{\sqrt{5}}{2} \tan θ$
$\tan θ = \frac{2}{\sqrt{5} t}$
$∴ \sin θ = \frac{2}{\sqrt{4 + 5 t^{2}}}$
From equation (iii) and equation (iv), we get -
$- 2 t = \frac{2}{\sqrt{4 + 5 t^{2}}}$
On squaring both sides, we get -
$t^{2} (4 + 5 t^{2}) = 1$
$5 t^{4} + 4 t^{2} = 1$

AP EAMCET-21.04.2019

Ellipse

120644 The value of ' $k$ ' so that the line $y = 2 x + k$ may touch the ellipse $3 x^{2} + 5 y^{2} = 15$ is

1

\pm \sqrt{23}

2

\pm \sqrt{13}

3

\pm \sqrt{33}

4

\pm \sqrt{32}

Explanation:

A Given,
Equation of line
$y = 2 x + k$
Ellipse $3 x^{2} + 5 y^{2} = 15$
From equation (i) and (ii),
$3 x^{2} + 5 (2 x + k)^{2} = 15$
$3 x^{2} + 20 x^{2} + 5 k^{2} + 20 xk = 15$
$23 x^{2} + 5 k^{2} + 20 kx - 15 = 0$
$b^{2} - 4 ac = 0$
$(20 k)^{2} - 4.23 (5 k^{2} - 15) = 0$
$- 60 k^{2} + 1380 = 0$
$k^{2} = 23$
$k = \pm \sqrt{23}$

AP EAMCET-17.09.2020

Ellipse

120645 If the tangent at the point $(1, 2)$ on the ellipse $3 x^{2} + 4 y^{2} = 19$ is also a tangent to the parabola $y^{2} - k x = 0$ , then $k =$

1

\frac{57}{16}

2

\frac{- 57}{64}

3

\frac{57}{64}

4

\frac{- 57}{16}

Explanation:

D Given,
$3 x^{2} + 4 y^{2} = 19$ , tangent point
$(1, 2)$
$3. (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$y^{2} - kx = 0$
$y^{2} - k (\frac{19 - 8 y}{3}) = 0$
$3 y^{2} + 8 ky - 19 k = 0$
$∴ D = (8 k)^{2} - 4 (3) (- 19 k) = 0$
$64 k^{2} = - 12 \times 19 k$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$
$3 x^{2} + 4 y^{2} = 19$ , tangent point $(1, 2)$
$3 . (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$

AP EAMCET-24.04.2018

Ellipse

120642 If the tangent at a point $P$ one the parabola $y^{2} =$ $3 x$ is parallel to the line $x + 2 y = 1$ and tangents at the point $Q$ and $R$ on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ are perpendicular to the line $x - y =$ 2, then the area of the triangle $P Q R$ is:

1

3 \sqrt{5}

2

\frac{9}{\sqrt{5}}

3

\frac{3}{2} \sqrt{5}

4

5 \sqrt{3}

Explanation:

A Given,
$y^{2} = 3 x$
Tangent $P (x_{1}, y_{1})$ is parallel to, $x + 2 y = 1$
Then slope at $P = \frac{- 1}{2}$
$2 y \frac{d y}{d x} = 3$
$\frac{d y}{d x} = \frac{3}{2 y} = \frac{- 1}{2}$
$y = - 3$
$∴ = (3, - 3)$
Similarly, $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $R = (\frac{- 4}{\sqrt{5}}, \frac{- 1}{\sqrt{5}})$
Area of $△ PQR = \frac{1}{2} | \begin{array}{ccc} 3 - 31 \\ \frac{4}{\sqrt{5}} \frac{1}{\sqrt{5}} 1 \\ \frac{- 4}{\sqrt{5}} \frac{- 1}{\sqrt{5}} 1 \end{array} |$
$= \frac{1}{2} [3 (\frac{2}{\sqrt{5}}) + 3 (\frac{8}{\sqrt{5}}) + 0]$
$= \frac{30}{2 \sqrt{5}}$
$= 3 \sqrt{5}$

JEE Main-29.01.2023

Ellipse

120643 Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , On comparing standard ellipse equation, $a^{2} = 19, b^{2} = 15$ , Equation of tangent of ellipse is -, $y = mx + \sqrt{19 m^{2} + 15}$ , $mx - y \pm \sqrt{19 m^{2} + 15} = 0$ , Perpendicular distance from centre of circle to tangent $=$ , 4, $| \frac{\pm \sqrt{19 m^{2} + 15}}{\sqrt{m^{2} + 1}} | = 4$ , $19 m^{2} + 15 = 16 m^{2} + 16$ , $m = \pm \frac{1}{\sqrt{3}}$ , $\tan θ = m = \frac{1}{\sqrt{3}}$ , $θ = \frac{π}{6}$ with $x$ -axis or $\frac{π}{3}$ with $y$ -axis, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{π}{3}$ ., 911. If the tangent drawn to the parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is the normal to the ellipse $4 x^{2} + 5 y^{2} = 20$ at $(\sqrt{5} \cos θ, 2 \sin θ)$ , then,

1

\frac{π}{4}

2

\frac{π}{3}

3

\frac{π}{12}

4

\frac{π}{6}

Explanation:

B Given,
$15 x^{2} + 19 y^{2} = 285$
$\frac{x^{2}}{19} + \frac{y^{2}}{15} = 1$
Ans: a
Exp: (A) : Given, tangent to parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is $y \cdot y_{1} = 2 (x + x_{1})$
$y \cdot 2 t = 2 (x + t^{2})$
$y t = x + t^{2}$
$y = \frac{x}{t} + t$
Normal to the ellipse, $4 x^{2} + 5 y^{2} = 20$
$\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1 at (\sqrt{5} \cos θ, 2 \sin θ)$
Slope of normal $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$ at $(\sqrt{5} \cos θ, 2 \sin θ)$
$∴$ Equation of line is -
$y - 2 \sin θ = \frac{\sqrt{5}}{2} \tan θ (x - \sqrt{5} \cos θ)$
$y = \frac{\sqrt{5}}{2} \tan θ \times x - \frac{\sin θ}{2}$
By comparing equation (i) and equation (ii), we get -
$t = - \frac{\sin θ}{2}$
$\sin θ = - 2 t$
$\frac{1}{t} = \frac{\sqrt{5}}{2} \tan θ$
$\tan θ = \frac{2}{\sqrt{5} t}$
$∴ \sin θ = \frac{2}{\sqrt{4 + 5 t^{2}}}$
From equation (iii) and equation (iv), we get -
$- 2 t = \frac{2}{\sqrt{4 + 5 t^{2}}}$
On squaring both sides, we get -
$t^{2} (4 + 5 t^{2}) = 1$
$5 t^{4} + 4 t^{2} = 1$

AP EAMCET-21.04.2019

Ellipse

120644 The value of ' $k$ ' so that the line $y = 2 x + k$ may touch the ellipse $3 x^{2} + 5 y^{2} = 15$ is

1

\pm \sqrt{23}

2

\pm \sqrt{13}

3

\pm \sqrt{33}

4

\pm \sqrt{32}

Explanation:

A Given,
Equation of line
$y = 2 x + k$
Ellipse $3 x^{2} + 5 y^{2} = 15$
From equation (i) and (ii),
$3 x^{2} + 5 (2 x + k)^{2} = 15$
$3 x^{2} + 20 x^{2} + 5 k^{2} + 20 xk = 15$
$23 x^{2} + 5 k^{2} + 20 kx - 15 = 0$
$b^{2} - 4 ac = 0$
$(20 k)^{2} - 4.23 (5 k^{2} - 15) = 0$
$- 60 k^{2} + 1380 = 0$
$k^{2} = 23$
$k = \pm \sqrt{23}$

AP EAMCET-17.09.2020

Ellipse

120645 If the tangent at the point $(1, 2)$ on the ellipse $3 x^{2} + 4 y^{2} = 19$ is also a tangent to the parabola $y^{2} - k x = 0$ , then $k =$

1

\frac{57}{16}

2

\frac{- 57}{64}

3

\frac{57}{64}

4

\frac{- 57}{16}

Explanation:

D Given,
$3 x^{2} + 4 y^{2} = 19$ , tangent point
$(1, 2)$
$3. (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$y^{2} - kx = 0$
$y^{2} - k (\frac{19 - 8 y}{3}) = 0$
$3 y^{2} + 8 ky - 19 k = 0$
$∴ D = (8 k)^{2} - 4 (3) (- 19 k) = 0$
$64 k^{2} = - 12 \times 19 k$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$
$3 x^{2} + 4 y^{2} = 19$ , tangent point $(1, 2)$
$3 . (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$

AP EAMCET-24.04.2018

Ellipse

120642 If the tangent at a point $P$ one the parabola $y^{2} =$ $3 x$ is parallel to the line $x + 2 y = 1$ and tangents at the point $Q$ and $R$ on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ are perpendicular to the line $x - y =$ 2, then the area of the triangle $P Q R$ is:

1

3 \sqrt{5}

2

\frac{9}{\sqrt{5}}

3

\frac{3}{2} \sqrt{5}

4

5 \sqrt{3}

Explanation:

A Given,
$y^{2} = 3 x$
Tangent $P (x_{1}, y_{1})$ is parallel to, $x + 2 y = 1$
Then slope at $P = \frac{- 1}{2}$
$2 y \frac{d y}{d x} = 3$
$\frac{d y}{d x} = \frac{3}{2 y} = \frac{- 1}{2}$
$y = - 3$
$∴ = (3, - 3)$
Similarly, $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $R = (\frac{- 4}{\sqrt{5}}, \frac{- 1}{\sqrt{5}})$
Area of $△ PQR = \frac{1}{2} | \begin{array}{ccc} 3 - 31 \\ \frac{4}{\sqrt{5}} \frac{1}{\sqrt{5}} 1 \\ \frac{- 4}{\sqrt{5}} \frac{- 1}{\sqrt{5}} 1 \end{array} |$
$= \frac{1}{2} [3 (\frac{2}{\sqrt{5}}) + 3 (\frac{8}{\sqrt{5}}) + 0]$
$= \frac{30}{2 \sqrt{5}}$
$= 3 \sqrt{5}$

JEE Main-29.01.2023

Ellipse

120643 Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , On comparing standard ellipse equation, $a^{2} = 19, b^{2} = 15$ , Equation of tangent of ellipse is -, $y = mx + \sqrt{19 m^{2} + 15}$ , $mx - y \pm \sqrt{19 m^{2} + 15} = 0$ , Perpendicular distance from centre of circle to tangent $=$ , 4, $| \frac{\pm \sqrt{19 m^{2} + 15}}{\sqrt{m^{2} + 1}} | = 4$ , $19 m^{2} + 15 = 16 m^{2} + 16$ , $m = \pm \frac{1}{\sqrt{3}}$ , $\tan θ = m = \frac{1}{\sqrt{3}}$ , $θ = \frac{π}{6}$ with $x$ -axis or $\frac{π}{3}$ with $y$ -axis, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{π}{3}$ ., 911. If the tangent drawn to the parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is the normal to the ellipse $4 x^{2} + 5 y^{2} = 20$ at $(\sqrt{5} \cos θ, 2 \sin θ)$ , then,

1

\frac{π}{4}

2

\frac{π}{3}

3

\frac{π}{12}

4

\frac{π}{6}

Explanation:

B Given,
$15 x^{2} + 19 y^{2} = 285$
$\frac{x^{2}}{19} + \frac{y^{2}}{15} = 1$
Ans: a
Exp: (A) : Given, tangent to parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is $y \cdot y_{1} = 2 (x + x_{1})$
$y \cdot 2 t = 2 (x + t^{2})$
$y t = x + t^{2}$
$y = \frac{x}{t} + t$
Normal to the ellipse, $4 x^{2} + 5 y^{2} = 20$
$\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1 at (\sqrt{5} \cos θ, 2 \sin θ)$
Slope of normal $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$ at $(\sqrt{5} \cos θ, 2 \sin θ)$
$∴$ Equation of line is -
$y - 2 \sin θ = \frac{\sqrt{5}}{2} \tan θ (x - \sqrt{5} \cos θ)$
$y = \frac{\sqrt{5}}{2} \tan θ \times x - \frac{\sin θ}{2}$
By comparing equation (i) and equation (ii), we get -
$t = - \frac{\sin θ}{2}$
$\sin θ = - 2 t$
$\frac{1}{t} = \frac{\sqrt{5}}{2} \tan θ$
$\tan θ = \frac{2}{\sqrt{5} t}$
$∴ \sin θ = \frac{2}{\sqrt{4 + 5 t^{2}}}$
From equation (iii) and equation (iv), we get -
$- 2 t = \frac{2}{\sqrt{4 + 5 t^{2}}}$
On squaring both sides, we get -
$t^{2} (4 + 5 t^{2}) = 1$
$5 t^{4} + 4 t^{2} = 1$

AP EAMCET-21.04.2019

Ellipse

120644 The value of ' $k$ ' so that the line $y = 2 x + k$ may touch the ellipse $3 x^{2} + 5 y^{2} = 15$ is

1

\pm \sqrt{23}

2

\pm \sqrt{13}

3

\pm \sqrt{33}

4

\pm \sqrt{32}

Explanation:

A Given,
Equation of line
$y = 2 x + k$
Ellipse $3 x^{2} + 5 y^{2} = 15$
From equation (i) and (ii),
$3 x^{2} + 5 (2 x + k)^{2} = 15$
$3 x^{2} + 20 x^{2} + 5 k^{2} + 20 xk = 15$
$23 x^{2} + 5 k^{2} + 20 kx - 15 = 0$
$b^{2} - 4 ac = 0$
$(20 k)^{2} - 4.23 (5 k^{2} - 15) = 0$
$- 60 k^{2} + 1380 = 0$
$k^{2} = 23$
$k = \pm \sqrt{23}$

AP EAMCET-17.09.2020

Ellipse

120645 If the tangent at the point $(1, 2)$ on the ellipse $3 x^{2} + 4 y^{2} = 19$ is also a tangent to the parabola $y^{2} - k x = 0$ , then $k =$

1

\frac{57}{16}

2

\frac{- 57}{64}

3

\frac{57}{64}

4

\frac{- 57}{16}

Explanation:

D Given,
$3 x^{2} + 4 y^{2} = 19$ , tangent point
$(1, 2)$
$3. (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$y^{2} - kx = 0$
$y^{2} - k (\frac{19 - 8 y}{3}) = 0$
$3 y^{2} + 8 ky - 19 k = 0$
$∴ D = (8 k)^{2} - 4 (3) (- 19 k) = 0$
$64 k^{2} = - 12 \times 19 k$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$
$3 x^{2} + 4 y^{2} = 19$ , tangent point $(1, 2)$
$3 . (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$

AP EAMCET-24.04.2018

Ellipse

120642 If the tangent at a point $P$ one the parabola $y^{2} =$ $3 x$ is parallel to the line $x + 2 y = 1$ and tangents at the point $Q$ and $R$ on the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{1} = 1$ are perpendicular to the line $x - y =$ 2, then the area of the triangle $P Q R$ is:

1

3 \sqrt{5}

2

\frac{9}{\sqrt{5}}

3

\frac{3}{2} \sqrt{5}

4

5 \sqrt{3}

Explanation:

A Given,
$y^{2} = 3 x$
Tangent $P (x_{1}, y_{1})$ is parallel to, $x + 2 y = 1$
Then slope at $P = \frac{- 1}{2}$
$2 y \frac{d y}{d x} = 3$
$\frac{d y}{d x} = \frac{3}{2 y} = \frac{- 1}{2}$
$y = - 3$
$∴ = (3, - 3)$
Similarly, $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $R = (\frac{- 4}{\sqrt{5}}, \frac{- 1}{\sqrt{5}})$
Area of $△ PQR = \frac{1}{2} | \begin{array}{ccc} 3 - 31 \\ \frac{4}{\sqrt{5}} \frac{1}{\sqrt{5}} 1 \\ \frac{- 4}{\sqrt{5}} \frac{- 1}{\sqrt{5}} 1 \end{array} |$
$= \frac{1}{2} [3 (\frac{2}{\sqrt{5}}) + 3 (\frac{8}{\sqrt{5}}) + 0]$
$= \frac{30}{2 \sqrt{5}}$
$= 3 \sqrt{5}$

JEE Main-29.01.2023

Ellipse

120643 Let a circle of radius 4 be concentric to the ellipse $15 x^{2} + 19 y^{2} = 285$ . Then the common tangents are inclined to the minor axis of the ellipse at the angle.
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-21.04.2019,Shift-I], So, Hence, Where, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , On comparing standard ellipse equation, $a^{2} = 19, b^{2} = 15$ , Equation of tangent of ellipse is -, $y = mx + \sqrt{19 m^{2} + 15}$ , $mx - y \pm \sqrt{19 m^{2} + 15} = 0$ , Perpendicular distance from centre of circle to tangent $=$ , 4, $| \frac{\pm \sqrt{19 m^{2} + 15}}{\sqrt{m^{2} + 1}} | = 4$ , $19 m^{2} + 15 = 16 m^{2} + 16$ , $m = \pm \frac{1}{\sqrt{3}}$ , $\tan θ = m = \frac{1}{\sqrt{3}}$ , $θ = \frac{π}{6}$ with $x$ -axis or $\frac{π}{3}$ with $y$ -axis, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{π}{3}$ ., 911. If the tangent drawn to the parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is the normal to the ellipse $4 x^{2} + 5 y^{2} = 20$ at $(\sqrt{5} \cos θ, 2 \sin θ)$ , then,

1

\frac{π}{4}

2

\frac{π}{3}

3

\frac{π}{12}

4

\frac{π}{6}

Explanation:

B Given,
$15 x^{2} + 19 y^{2} = 285$
$\frac{x^{2}}{19} + \frac{y^{2}}{15} = 1$
Ans: a
Exp: (A) : Given, tangent to parabola $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is $y \cdot y_{1} = 2 (x + x_{1})$
$y \cdot 2 t = 2 (x + t^{2})$
$y t = x + t^{2}$
$y = \frac{x}{t} + t$
Normal to the ellipse, $4 x^{2} + 5 y^{2} = 20$
$\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1 at (\sqrt{5} \cos θ, 2 \sin θ)$
Slope of normal $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$ at $(\sqrt{5} \cos θ, 2 \sin θ)$
$∴$ Equation of line is -
$y - 2 \sin θ = \frac{\sqrt{5}}{2} \tan θ (x - \sqrt{5} \cos θ)$
$y = \frac{\sqrt{5}}{2} \tan θ \times x - \frac{\sin θ}{2}$
By comparing equation (i) and equation (ii), we get -
$t = - \frac{\sin θ}{2}$
$\sin θ = - 2 t$
$\frac{1}{t} = \frac{\sqrt{5}}{2} \tan θ$
$\tan θ = \frac{2}{\sqrt{5} t}$
$∴ \sin θ = \frac{2}{\sqrt{4 + 5 t^{2}}}$
From equation (iii) and equation (iv), we get -
$- 2 t = \frac{2}{\sqrt{4 + 5 t^{2}}}$
On squaring both sides, we get -
$t^{2} (4 + 5 t^{2}) = 1$
$5 t^{4} + 4 t^{2} = 1$

AP EAMCET-21.04.2019

Ellipse

120644 The value of ' $k$ ' so that the line $y = 2 x + k$ may touch the ellipse $3 x^{2} + 5 y^{2} = 15$ is

1

\pm \sqrt{23}

2

\pm \sqrt{13}

3

\pm \sqrt{33}

4

\pm \sqrt{32}

Explanation:

A Given,
Equation of line
$y = 2 x + k$
Ellipse $3 x^{2} + 5 y^{2} = 15$
From equation (i) and (ii),
$3 x^{2} + 5 (2 x + k)^{2} = 15$
$3 x^{2} + 20 x^{2} + 5 k^{2} + 20 xk = 15$
$23 x^{2} + 5 k^{2} + 20 kx - 15 = 0$
$b^{2} - 4 ac = 0$
$(20 k)^{2} - 4.23 (5 k^{2} - 15) = 0$
$- 60 k^{2} + 1380 = 0$
$k^{2} = 23$
$k = \pm \sqrt{23}$

AP EAMCET-17.09.2020

Ellipse

120645 If the tangent at the point $(1, 2)$ on the ellipse $3 x^{2} + 4 y^{2} = 19$ is also a tangent to the parabola $y^{2} - k x = 0$ , then $k =$

1

\frac{57}{16}

2

\frac{- 57}{64}

3

\frac{57}{64}

4

\frac{- 57}{16}

Explanation:

D Given,
$3 x^{2} + 4 y^{2} = 19$ , tangent point
$(1, 2)$
$3. (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$y^{2} - kx = 0$
$y^{2} - k (\frac{19 - 8 y}{3}) = 0$
$3 y^{2} + 8 ky - 19 k = 0$
$∴ D = (8 k)^{2} - 4 (3) (- 19 k) = 0$
$64 k^{2} = - 12 \times 19 k$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$
$3 x^{2} + 4 y^{2} = 19$ , tangent point $(1, 2)$
$3 . (x) 1 + 4 (y) (2) = 19$
$3 x + 8 y = 19$
$k^{2} = - \frac{12 \times 19 k}{64}$
$k = \frac{- 57}{16}$

AP EAMCET-24.04.2018

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