Conic Sections

Let $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1,$ where

The correct option is:

*a*>*b*> 0, be a hyperbola in the*x*y-plane whose conjugate axis*LM*subtends an angle of 60° at one of its vertices*N*. Let the area of the triangle*LMN*be $4\sqrt{3}$.LIST–I |
LIST–II |
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P. | The length of the conjugate axis of H is |
1. | 8 |

Q. | The eccentricity of H is |
2. | $\frac{4}{\sqrt{3}}$ |

R | The distance between the foci of H is |
3. | $\frac{2}{\sqrt{3}}$ |

S. | The length of the latus rectum of H is |
4. | 4 |

The correct option is:

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JEE Advanced 2018

Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively. | ||

Column-1 |
Column-2 |
Column-3 |

(I) x^{2} + y^{2} = a^{2} |
(i) my = m^{2}x + a |
(P) $\left(\frac{\mathrm{a}}{{\mathrm{m}}^{2}},\frac{2\mathrm{a}}{2}\right)$ |

(II) x^{2} + a^{2}y^{2} = a^{2} |
(ii) $\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{{\mathrm{m}}^{2}+1}$ | (Q) $\left(\frac{-\mathrm{ma}}{\sqrt{{\mathrm{m}}^{2}+1}},\frac{\mathrm{a}}{\sqrt{{\mathrm{m}}^{2}+1}}\right)$ |

(III) y^{2} = 4ax |
(iii) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}$ | (R) $\left(\frac{-{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}}\right)$ |

(IV) x^{2} – a^{2}y^{2} = a^{2} |
(iv) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}$ | (S) $\left(\frac{-{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}},\frac{-1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}}\right)$ |

If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8,16), then which of the following options is the only

**CORRECT**combination ?

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JEE Advanced 2017

Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively. | ||

Column-1 |
Column-2 |
Column-3 |

(I) x^{2} + y^{2} = a^{2} |
(i) my = m^{2}x + a |
(P) $\left(\frac{\mathrm{a}}{{\mathrm{m}}^{2}},\frac{2\mathrm{a}}{2}\right)$ |

(II) x^{2} + a^{2}y^{2} = a^{2} |
(ii) $\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{{\mathrm{m}}^{2}+1}$ | (Q) $\left(\frac{-\mathrm{ma}}{\sqrt{{\mathrm{m}}^{2}+1}},\frac{\mathrm{a}}{\sqrt{{\mathrm{m}}^{2}+1}}\right)$ |

(III) y^{2} = 4ax |
(iii) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}$ | (R) $\left(\frac{-{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}}\right)$ |

(IV) x^{2} – a^{2}y^{2} = a^{2} |
(iv) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}$ | (S) $\left(\frac{-{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}},\frac{-1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}}\right)$ |

For $\mathrm{a}=\sqrt{2}$ , if a tangent is drawn to a suitable conic (Column 1) at the point of contact (–1,1), then which of the following options is the only

**CORRECT**combination for obtaining its equation ?

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JEE Advanced 2017

Column 1, 2 and 3 contain conics, equation of tangents to the conics and points of contact, respectively. | ||

Column-1 |
Column-2 |
Column-3 |

(I) x^{2} + y^{2} = a^{2} |
(i) my = m^{2}x + a |
(P) $\left(\frac{\mathrm{a}}{{\mathrm{m}}^{2}},\frac{2\mathrm{a}}{2}\right)$ |

(II) x^{2} + a^{2}y^{2} = a^{2} |
(ii) $\mathrm{y}=\mathrm{mx}+\mathrm{a}\sqrt{{\mathrm{m}}^{2}+1}$ | (Q) $\left(\frac{-\mathrm{ma}}{\sqrt{{\mathrm{m}}^{2}+1}},\frac{\mathrm{a}}{\sqrt{{\mathrm{m}}^{2}+1}}\right)$ |

(III) y^{2} = 4ax |
(iii) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}$ | (R) $\left(\frac{-{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}},\frac{1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}}\right)$ |

(IV) x^{2} – a^{2}y^{2} = a^{2} |
(iv) $\mathrm{y}=\mathrm{mx}+\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}+1}$ | (S) $\left(\frac{-{\mathrm{a}}^{2}\mathrm{m}}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}},\frac{-1}{\sqrt{{\mathrm{a}}^{2}{\mathrm{m}}^{2}-1}}\right)$ |

The tangent to a suitable conic (Column 1) at $\left(\sqrt{3},\frac{1}{2}\right)$ is found to be $\sqrt{3}\mathrm{x}+2\mathrm{y}=4$, then which of the following options is the only

**CORRECT**combination?

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JEE Advanced 2017

PARAGRAPH 2Let F
_{1}(x_{1}, 0) and F_{2}(x_{2}, 0), for x_{1} < 0 and x_{2} > 0, be the foci of the ellips $\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{8}=1.$Suppose a parabola having vertex at the origin and focus at F _{2} intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. |

If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the

*x*-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF

_{1}NF

_{2}is

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JEE Advanced 2016

PARAGRAPH 2Let F
_{1}(x_{1}, 0) and F_{2}(x_{2}, 0), for x_{1} < 0 and x_{2} > 0, be the foci of the ellips $\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{8}=1.$Suppose a parabola having vertex at the origin and focus at F _{2} intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. |

The orthocentre of the triangle F

_{1}MN is

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JEE Advanced 2016

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JEE Advanced 2012

Choose the option corresponding to the correct matching of elements from List-I and List-II

List-I |
List-II |
||

P. | The coordinates of the mid-point of the line segment formed by joining the point of contact of vertical tangents to 9x^{2} + 4y^{2} – 18x –16y –11 = 0 |
1. | (3, 4) |

Q. | The function $f\left(x\right)={\int}_{0}^{x\left(3-{x}^{2}\right)}{t}^{2}\xb7{e}^{\mathrm{sin}t}\mathit{d}t$ increases in the open interval | 2. | (2, 3) |

R. | The function $g\left(x\right)=\left\{\begin{array}{cc}{\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right),& 2<x\le 3\\ {\mathrm{tan}}^{-1}\left(\mathrm{tan}x\right)& 3<x<4\end{array}\right.\begin{array}{}\\ \end{array}$decreases in the interval | 3. | (1, ∞) |

S. | Let the normal at the point whose eccentric angle is $\theta ={\mathrm{cot}}^{-1}\frac{4}{3}$on the hyperbola x^{2 }– 4y^{2} – 2x – 63 = 0 intersects the coordinate axes at A and B. Coordinates of the mid-point of line segment AB is |
4. | (1, 2) |

5. | (–∞, –1) | ||

6. | (–1, 1) | ||

7. | $\left(\frac{27}{4},\frac{81}{10}\right)$ |

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JEE Advanced 0

Columns 1, 2 and 3 contains equations of circles, equations of secants and length of corresponding chords respectively. Answer the following question based on it

If a secant passes through a point (1, 0) then which of the following is correct?

Column 1 |
Column 2 |
Column 3 |
|||

(I) | x^{2} + y^{2} – 4x + 3 = 0 |
(i) | x – y + 1 = 0 |
(P) | $\sqrt{14}$ units |

(II) | x^{2} + y^{2} + 4x + 2 = 0 |
(ii) | x + 3y + 9 = 0 |
(Q) | $\frac{8}{\sqrt{5}}$ units |

(III) | x^{2} + y^{2} – 4 = 0 |
(iii) | 3x – y – 3 = 0 |
(R) | $\sqrt{6}$ units |

(IV) | x^{2} + y^{2} + 6x + 2y + 6 = 0 |
(iv) | 2x + y – 2 = 0 |
(S) | $\frac{2}{\sqrt{5}}$ units |

If a secant passes through a point (1, 0) then which of the following is correct?

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JEE Advanced 0

Columns 1, 2 and 3 contains equations of circles, equations of secants and length of corresponding chords respectively. Answer the following question based on it

If a secant of the suitable circle passes through a point (0, 2) then which of the following combination is correct?

Column 1 |
Column 2 |
Column 3 |
|||

(I) | x^{2} + y^{2} – 4x + 3 = 0 |
(i) | x – y + 1 = 0 |
(P) | $\sqrt{14}$ units |

(II) | x^{2} + y^{2} + 4x + 2 = 0 |
(ii) | x + 3y + 9 = 0 |
(Q) | $\frac{8}{\sqrt{5}}$ units |

(III) | x^{2} + y^{2} – 4 = 0 |
(iii) | 3x – y – 3 = 0 |
(R) | $\sqrt{6}$ units |

(IV) | x^{2} + y^{2} + 6x + 2y + 6 = 0 |
(iv) | 2x + y – 2 = 0 |
(S) | $\frac{2}{\sqrt{5}}$ units |

If a secant of the suitable circle passes through a point (0, 2) then which of the following combination is correct?

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JEE Advanced 0

Columns 1, 2 and 3 contains equations of circles, equations of secants and length of corresponding chords respectively. Answer the following question based on it

A suitable circle passes through a point (–3, 1) then which of the following is the correct combination?

Column 1 |
Column 2 |
Column 3 |
|||

(I) | x^{2} + y^{2} – 4x + 3 = 0 |
(i) | x – y + 1 = 0 |
(P) | $\sqrt{14}$ units |

(II) | x^{2} + y^{2} + 4x + 2 = 0 |
(ii) | x + 3y + 9 = 0 |
(Q) | $\frac{8}{\sqrt{5}}$ units |

(III) | x^{2} + y^{2} – 4 = 0 |
(iii) | 3x – y – 3 = 0 |
(R) | $\sqrt{6}$ units |

(IV) | x^{2} + y^{2} + 6x + 2y + 6 = 0 |
(iv) | 2x + y – 2 = 0 |
(S) | $\frac{2}{\sqrt{5}}$ units |

A suitable circle passes through a point (–3, 1) then which of the following is the correct combination?

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JEE Advanced 0

Observe the following columns :

Column I |
Column II |
||

(A) | The minimum and maximum distance of a point (2, 6) from the ellipse 9x^{2} + 8y^{2} – 36x – 16y – 28 = 0 are L and G, then |
(P) | L + G = 10 |

(B) | The minimum and maximum distance of a point (1, 2) from the ellipse 4x^{2} + 9y^{2} + 8x – 36y + 4 = 0 are L and G, then |
(Q) | L + G = 6 |

(C) | The minimum and maximum distance of a point $\left(\frac{9}{5},\frac{12}{5}\right)$ from the ellipse 4(3x + 4y)^{2} + 9(4x – 3y)^{2} = 900 are L and G, then |
(R) | G – L = 6 |

(S) | G – L = 4 | ||

(T) | L^{G} + G^{L} = 6 |

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JEE Advanced 0