SATHEE: Hyperbola (Lecture-02)

Hyperbola (Lecture-02)

Practice Problems

1. Let $C$ be a curve which is locus of the point of the intersection of lines $x = 2 + m$ and $my = 4 - m$ . A circle $(x - 2)^{2} + (y + 1)^{2} = 25$ intersects the curve $C$ at four points $P, Q, R$ and $S$ . If $O$ is the centre of curve ’ $C$ ’ than ${OP}^{2} + {OQ}^{2} + {OR}^{2} + {OS}^{2}$ is

(a) 25

(b) 50

(d) 100

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Solution

$\begin{aligned} x - 2 = m \\ y + 1 = \frac{4}{m} \\ ∴ (x - 2) (y + 1) = 4 \\ XY = 4, where X = x - 2 and Y = y + 1 \\ x - 2)^{2} + (y + 1)^{2} = 25 \\ X^{2} + Y^{2} = 25 \end{aligned}$

$and (x - 2)^{2} + (y + 1)^{2} = 25$

Curve $C$ and circle both are concentric.

$∴ {OP}^{2} + {OQ}^{2} + {OR}^{2} + {OS}^{2} = 4 r^{2}$

$\begin{aligned} = 4 (25) \\ = 100 \end{aligned}$

Answer (d)

2. If $(5, 12)$ and $(24, 7)$ are the foci of a hyperbola passing through the origin, then

(a) $e = \frac{\sqrt{386}}{12}$

(b) $e = \frac{\sqrt{386}}{36}$

(d) $LR = \frac{121}{9}$

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Solution

$\begin{aligned} | {PS}_{1} - S_{2} | = 2 a P (0, 0) \\ \sqrt{(24 - 0)^{2} + (7 - 0)^{2}} - \sqrt{(5 - 0)^{2} + (12 - 0)^{2}} = 2 a \\ 25 - 13 = 2 a \\ ∴ a = 6 \\ 2 ae = \sqrt{(24 - 5)^{2} + (7 - 12)^{2}} = \sqrt{386} \\ e = \frac{\sqrt{386}}{12} \\ LR = \frac{2 b^{2}}{a} = \frac{2 a^{2} (e^{2} - 1)}{a} = 2 \times 6 (\frac{386}{144} - 1) = \frac{121}{6} \end{aligned}$

Answer (a,c)

3. If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $xy = c^{2}$ in four points $P (x_{1}, y_{1}), Q (x_{2}, y_{2}), R (x_{3}, y_{3})$ , $S (x_{4}, y_{4})$ then

(a) $x_{1} + x_{2} + x_{3} + x_{4} = 0$

(b) $y_{1} + y_{2} + y_{3} + y_{4} = 0$

(d) $y_{1} y_{2} y_{3} y_{4} = c^{4}$

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Solution

solving $xy = c^{2}$ and $x^{2} + y^{2} = a^{2}$ , we get

$\begin{matrix} x^{2} + \frac{c^{4}}{x^{2}} = a^{2} \\ x^{4} - a^{2} x^{2} + c^{4} = 0 \\ x_{1} + x_{2} + x_{3} + x_{4} = 0 \\ x_{1} \cdot x_{2} \cdot x_{3} \cdot x_{4} = c^{4} \\ similarly \frac{c^{4}}{y^{2}} + y^{2} = a^{2} \\ y^{4} - a^{2} y^{2} + c^{4} = 0 \\ y_{1} + y_{2} + y_{3} + y_{4} = 0 \\ y_{1} \cdot y_{2} \cdot y_{4} \cdot y_{4} = c^{4} \end{matrix}$

Answer a, b, c, d

Linked comprehension Type (For problem 4-6)

The vertices of $△ ABC$ lie on a rectangular hyperbola such that the orthocenter of the triangle is $(3, 2)$ and the asymptotes of the rectangular hyperbola are parallel to the wordinate axes. The two perpendicular tangents of the hyperbola intersect at the point $(1, 1)$ .

4. The equation of rectangular hyperbola is

(a) $x y = x + y - 1$

(b) $x y = x + y + 1$

(d) $x y = x + y$

5. The equation of asymptotes is

(a) $(x - 1) (y + 1) = 0$

(b) $(x + 1) (y - 1) = 0$

(d) $(x + 1) (y + 1) = 0$

6. Number of real tangents that can be drawn from the point $(1, 1)$ to the rectangular hyperbola is

(a) 4

(b) 1

(d) 2

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Solution

Perpendicular tangents intersect at the centre of rectangular hyperbola.

Hence centre is $(1, 1)$

Equation of asymptotes are $(x - 1) (y - 1) = 0$

Equation of hyperbola is $x y - x - y + 1 + λ = 0$

If panes through $(3, 1)$ , hence $λ = - 2$

Equation of hyperbola is $xy - x - y - 1 = 0$

Ans b
Ans c
From centre of hyperbola we can drow two real tangents.

Answer (d)

7. True/ False

S1: Number of points from where perpendicular tangents can be drown to the hyperbola $16 x^{2} - 9 y^{2} = 144$ is infinite.

S2: If distance between two parallel tangents drawn to the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{49} = 1$ is 2 then

their slopes is equal to $\pm \frac{5}{2}$ .

S3: If through the point $(5, 0)$ chords are drawn to the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1$ . Then locus of their middle points is also a hyperbola whose length of latus rectum is same as given hyperbola $9 x^{2} - 25 y^{2} = 225$ .

S4: If the line $y = m x + \sqrt{a^{2} m^{2} - b^{2}}$ touches the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at the point $(asec θ, b \tan θ)$ then $θ = \sin^{- 1} (\frac{b}{am})$ .

(a) TTFF

(b) FTTF

(d) TFTF

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Solution

S1: If $b^{2} > a^{2}$ , the radius of director circle is imaginary. Hence there is no pair of tangents at right angle can be drawn to the curve.

S2: $y = m x \pm \sqrt{9 m^{2} - 49}$

Now $| \frac{2 \sqrt{9 m^{2} - 49}}{\sqrt{1 + m^{2}}} | = 2 \Rightarrow 9 m^{2} - 49 = 1 + m^{2}$

$\Rightarrow 8 m^{2} = 50$

$\Rightarrow m = \pm \frac{5}{2}$

S3: Let mid point be $(h, k)$

Equation of chord whose mid point is given is

$T = S_{1}$

$\frac{xh}{25} - \frac{yk}{9} - 1 = \frac{b^{2}}{25} - \frac{k^{2}}{9} - 1$

This passes through $(5, 0) \frac{5 h}{25} - 0 - 1 = \frac{h^{2}}{25} - \frac{K^{2}}{9} - 1$

$45 h = 9 h^{2} - 25 k^{2}$

Locus of (h,k) is $9 x^{2} - 45 x - 25 y^{2} = 0$

$\begin{aligned} 9 (x^{2} - 5 x + \frac{25}{4} - \frac{25}{4}) - 25 y^{2} = 0 \\ 9 (x - 5 / 2)^{2} - 25 y^{2} = \frac{225}{4} \end{aligned}$

$\frac{(x - 5 / 2)^{2}}{(\frac{25}{4})} - \frac{y^{2}}{(\frac{9}{4})} = 1$

Length of latus rectum in not same.

S4: Since $(asec θ, b \tan θ)$ lies on $y = m x + \sqrt{a^{2} m^{2} - b^{2}}$

$∴ btan θ = amsec θ + \sqrt{a^{2} m^{2} - b^{2}}$

$(b \tan θ - a \sec θ)^{2} = a^{2} m^{2} - b^{2}$

$a^{2} m^{2} \sin^{2} θ -$ eabmsin $θ + b^{2} = 0 (∵ \cos θ \neq 0)$

$(am \sin θ - b)^{2} = 0$

$∴ \sin θ = \frac{b}{am}$

Hence $θ = \sin^{- 1} (\frac{b}{am})$

Answer (c)

8. Match the column

	Column I		Column II
(a)	The area of the triangle that a tangent at a point of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ makes with its asymptotes is	(p)	12
(b)	If the line $y = 3 x + λ$ touches the curve $9 x^{2} - 5 y^{2} = 45$ , then $\| λ \|$ is	(q)	6
(c)	If the chord $x \cos α + y \sin α = p$ of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{18} = 1$ subtends a right angle at the centre, then the diameter of the circle, concentric with the hyperbola, to which the given chord is a tangent is	(r)	24
(d)	If $λ$ be the length of the latus rectum of the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , then $3 λ$ is equal to	(s)	32

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Solution

(a) Equation of tangent at $(a, 0)$ is $x = a$

asymptotes $y = \frac{b}{a} x$ & $y = - \frac{b}{a} x$

$Area = 2 \times \frac{1}{2} \times a \times b = 4 \times 3 = 12 a$

(b) $y = 3 x + λ$ touches the curve $9 x^{2} - 5 y^{2} = 45$

$\begin{aligned} then 9 x^{2} - 5 (3 x + λ)^{2} = 45 \\ 36 x^{2} + 30 λ x + 5 λ^{2} + 45 = 0 has equal roots \\ ∴ 900 λ^{2} - 4 \times 36 (5 λ^{2} + 45) = 0 \\ λ^{2} = 36 \Rightarrow λ = \pm 6 \\ Hence | λ | = 6 \end{aligned}$

(c) The combined equation $OA$ and $OB$ we get when we make a homogeneous equation with the help of line and hyperbola

i.e. $18 x^{2} - 16 y^{2} - 288 {(\frac{x \cos α + y \sin α}{p})}^{2} = 0$

$9 p^{2} x^{2} - 9 p^{2} y^{2} - 144 (x^{2} \cos^{2} α + y^{2} \sin^{2} α + 2 x y \sin α \cos α) = 0$

$(9 p^{2} - 144 \cos^{2} α) x^{2} - 288 \sin α \cos α \cdot x y + (- 144 \sin^{2} α - 8 p^{2}) y^{2} = 0$

Lines are perpendicular $\Rightarrow 9 p^{2} - 144 \cos^{2} α - 8 p^{2} - 144 \sin^{2} α = 0$

$\begin{aligned} p^{2} = 144 \\ p = \pm 12 \end{aligned}$

Radius of circle $= 12$

$∴$ Diameter of circle $= 24$

(d) $16 x^{2} + 32 x - 9 y^{2} + 36 y = 164$

$16 (x + 1)^{2} - 9 (y - 2)^{2} = 144$

$\frac{(x + 1)^{2}}{9} - \frac{(y - 2)^{2}}{16} = 1$

length of latus rectum $= 2 \times \frac{16}{3} = \frac{32}{3}$

$∴ 3 λ = 32$

$(d) \to (s)$

Exercis - 9

1. For a given non zero value of $m$ each of the lines $\frac{x}{a} - \frac{y}{b} = m$ and $\frac{x}{a} + \frac{y}{b} = m$ meets the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at a point. Sum of the ordinates of these points, is

(a) $\frac{a (1 + m^{2})}{m}$

(b) $\frac{a + b}{2 m}$

(d) 0

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Answer: d

2. For which of the hyperbola, we can have more than one pair of perpendicular tangents?

(a) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = - 1$

(b) $\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1$

(d) $x y = 9$

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Answer: a

3. From point $(2, 2)$ tangents are drawn to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ then point of contact lie in

(a) I & IV quadrants

(b) II & III quadrants

(d) II & III quadrants

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Answer: c

4. If $(5, 12)$ and $(24, 7)$ are the foci of a conic passing through the origin then the eccentricity of conic is

(a) $\frac{\sqrt{386}}{13}$

(b) $\frac{\sqrt{386}}{12}$

(d) $\frac{\sqrt{386}}{25}$

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Answer: c

5. For the hyperbola $9 x^{2} - 16 y^{2} - 18 x + 32 y - 151 = 0$

(a) directrix is $x = \frac{21}{5}$

(b) latus rectum $= \frac{9}{2}$

(d) foci are $(6, 1)$ and $(- 4, 1)$

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Answer: d

6. Assertion / Reasoning

Statement - 1 If a circle $S = 0$ intersects a hyperbola $x y = 4$ at four points. Three of then are $(2, 2), (4, 1)$ and $(6, \frac{2}{3})$ then coordinates of the fourth point are $(\frac{1}{4}, 16)$ .

Statement - 2 If a circle $S = 0$ intersects a hyperbola $x y = c^{2}$ at $t_{1}, t_{2}, t_{3}, t_{4}$ then $t_{1} \cdot t_{2} \cdot t_{3} \cdot t_{4} = 1$

(a) Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement -1

(b) Statement -1 is True, Statement - 2 is True, Statement -2 is NOT a correct explanation for Statement - 1

(d) Statement - 1 is False, statement 2 is True.

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Answer: d

True/ false

7. S1: Centre of the hyperbola $x^{2} - 4 y^{2} - 4 x + 8 y + 4 = 0$ is $(2, 1)$

S2 : If eccentricity of hyperbola $x (y - 1) = 2$ is $\sqrt{2}$ then eccentricity of its conjugate hyperbola is 2 .

S3 : From point $(2, 2)$ tangents are drawn to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ , then point of contact lie in I & IV quadrants.

S4 : Product of the length of perpendiculars drawn from any foci of the hyperbola $x^{2} - 4 y^{2} -$ $4 x + 8 y + 4 = 0$ to its asymptotes is 4 .

(a) TFTT

(b) TFFT

(d) TTTT

Comprehension

For the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ the normal at $P$ meets the transverse axis $A^{'}$ in $D$ and the conjugate axis ${BB}^{'}$ in $E$ and $CF$ be perpendicular to the normal from the centre.

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Answer: b

8. $PF \times PD = k (CB)^{2}$ , then $k =$

(a) $\frac{1}{2}$

(b) 1

(d) 3

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Answer: b

9. $PF \times PE$ equal to

A program to give wings to girl students

(a) $CA \times CB$

(b) $(CB)^{2}$

(d) $(CF)^{2}$

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Answer: c

10. Locus of middle point of $D$ and $E$ is a hyperbola of eccentricity

(a) $\sqrt{e^{2} - 1}$

(b) $\frac{1}{\sqrt{e^{2} - 1}}$

(d) $\frac{e}{\sqrt{e^{2} - 1}}$

Show Answer

Answer: d

Hyperbola

Hyperbola (Lecture-02)

Practice Problems

Linked comprehension Type (For problem 4-6)

Exercis - 9

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Hyperbola

Hyperbola (Lecture-02)

Practice Problems

Linked comprehension Type (For problem 4-6)

Exercis - 9

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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