Hyperbola (Lecture-02)
Practice Problems
1. Let $\mathrm{C}$ be a curve which is locus of the point of the intersection of lines $\mathrm{x}=2+\mathrm{m}$ and $\mathrm{my}=4-\mathrm{m}$. 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 ’ $\mathrm{C}$ ’ than $\mathrm{OP}^{2}+\mathrm{OQ}^{2}+\mathrm{OR}^{2}+\mathrm{OS}^{2}$ is
(a) 25
(b) 50
(c) $\dfrac{25}{2}$
(d) 100
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Solution
$\begin{aligned} & \mathrm{x}-2=\mathrm{m} \\ & \mathrm{y}+1=\dfrac{4}{\mathrm{~m}} \\ & \therefore(\mathrm{x}-2)(\mathrm{y}+1)=4 \\ & \mathrm{XY}=4, \text { where } \mathrm{X}=\mathrm{x}-2 \text { and } \mathrm{Y}=\mathrm{y}+1 \\ & \mathrm{x}-2)^{2}+(\mathrm{y}+1)^{2}=25 \\ & \mathrm{X}^{2}+\mathrm{Y}^{2}=25 \end{aligned}$
$\text { and }(x-2)^{2}+(y+1)^{2}=25$
Curve $\mathrm{C}$ and circle both are concentric.
$\therefore \mathrm{OP}^{2}+\mathrm{OQ}^{2}+\mathrm{OR}^{2}+\mathrm{OS}^{2}=4 \mathrm{r}^{2}$
$\begin{aligned} & =4(25) \\ & =100 \end{aligned}$
Answer (d)
2. If $(5,12)$ and $(24,7)$ are the foci of a hyperbola passing through the origin, then
(a) $\mathrm{e}=\dfrac{\sqrt{386}}{12}$
(b) $\mathrm{e}=\dfrac{\sqrt{386}}{36}$
(c) $\mathrm{LR}=\dfrac{121}{6}$
(d) $\mathrm{LR}=\dfrac{121}{9}$
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Solution
$\begin{aligned} & \left|\mathrm{PS} _{1}-\mathrm{S} _{2}\right|=2 \mathrm{a} \quad \mathrm{P}(0,0) \\ & \sqrt{(24-0)^{2}+(7-0)^{2}}-\sqrt{(5-0)^{2}+(12-0)^{2}}=2 \mathrm{a} \\ & 25-13=2 \mathrm{a} \\ & \quad \therefore \mathrm{a}=6 \\ & 2 \mathrm{ae}=\sqrt{(24-5)^{2}+(7-12)^{2}}=\sqrt{386} \\ & \quad \mathrm{e}=\dfrac{\sqrt{386}}{12} \\ & \mathrm{LR}=\dfrac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\dfrac{2 \mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)}{\mathrm{a}}=2 \times 6\left(\dfrac{386}{144}-1\right)=\dfrac{121}{6} \end{aligned}$
Answer (a,c)
3. If the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ intersects the hyperbola $\mathrm{xy}=\mathrm{c}^{2}$ in four points $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right), \mathrm{Q}\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right), \mathrm{R}\left(\mathrm{x} _{3}, \mathrm{y} _{3}\right)$, $\mathrm{S}\left(\mathrm{x} _{4}, \mathrm{y} _{4}\right)$ then
(a) $\mathrm{x} _{1}+\mathrm{x} _{2}+\mathrm{x} _{3}+\mathrm{x} _{4}=0$
(b) $\mathrm{y} _{1}+\mathrm{y} _{2}+\mathrm{y} _{3}+\mathrm{y} _{4}=0$
(c) $\mathrm{x} _{1} \mathrm{x} _{2} \mathrm{x} _{3} \mathrm{x} _{4}=\mathrm{c}^{4}$
(d) $\mathrm{y} _{1} \mathrm{y} _{2} \mathrm{y} _{3} \mathrm{y} _{4}=\mathrm{c}^{4}$
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Solution
solving $\mathrm{xy}=\mathrm{c}^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$, we get
$\begin{gathered} \mathrm{x}^{2}+\dfrac{\mathrm{c}^{4}}{\mathrm{x}^{2}}=\mathrm{a}^{2} \\ \mathrm{x}^{4}-\mathrm{a}^{2} \mathrm{x}^{2}+\mathrm{c}^{4}=0 \\ \mathrm{x} _{1}+\mathrm{x} _{2}+\mathrm{x} _{3}+\mathrm{x} _{4}=0 \\ \mathrm{x} _{1} \cdot \mathrm{x} _{2} \cdot \mathrm{x} _{3} \cdot \mathrm{x} _{4}=\mathrm{c}^{4} \\ \text { similarly } \dfrac{\mathrm{c}^{4}}{\mathrm{y}^{2}}+\mathrm{y}^{2}=\mathrm{a}^{2} \\ \mathrm{y}^{4}-\mathrm{a}^{2} \mathrm{y}^{2}+\mathrm{c}^{4}=0 \\ \mathrm{y} _{1}+\mathrm{y} _{2}+\mathrm{y} _{3}+\mathrm{y} _{4}=0 \\ \mathrm{y} _{1} \cdot \mathrm{y} _{2} \cdot \mathrm{y} _{4} \cdot \mathrm{y} _{4}=\mathrm{c}^{4} \\ \end{gathered}$
Answer a, b, c, d
Linked comprehension Type (For problem 4-6)
The vertices of $\triangle \mathrm{ABC}$ 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) $x y=x+y+1$
(c) $2 x y=x+y$
(d) $x y=x+y$
5. The equation of asymptotes is
(a) $(x-1)(y+1)=0$
(b) $(x+1)(y-1)=0$
(c) $(\mathrm{x}-1)(\mathrm{y}-1)=0$
(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
(c) 0
(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 $x y-x-y+1+\lambda=0$
If panes through $(3,1)$, hence $\lambda=-2$
Equation of hyperbola is $\mathrm{xy}-\mathrm{x}-\mathrm{y}-1=0$
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Ans b
-
Ans c
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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 $\dfrac{x^{2}}{9}-\dfrac{y^{2}}{49}=1$ is 2 then
their slopes is equal to $\pm \dfrac{5}{2}$.
S3: If through the point $(5,0)$ chords are drawn to the hyperbola $\dfrac{x^{2}}{25}-\dfrac{y^{2}}{9}=1$. Then locus of their middle points is also a hyperbola whose length of latus rectum is same as given hyperbola $9 \mathrm{x}^{2}-25 \mathrm{y}^{2}=225$.
S4: If the line $y=m x+\sqrt{a^{2} m^{2}-b^{2}}$ touches the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ at the point $(\operatorname{asec} \theta, b \tan \theta)$ then $\theta=\sin ^{-1}\left(\dfrac{\mathrm{b}}{\mathrm{am}}\right)$.
(a) TTFF
(b) FTTF
(c) FTFT
(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 $\left|\dfrac{2 \sqrt{9 \mathrm{~m}^{2}-49}}{\sqrt{1+\mathrm{m}^{2}}}\right|=2 \Rightarrow 9 \mathrm{~m}^{2}-49=1+\mathrm{m}^{2}$
$\Rightarrow \quad 8 \mathrm{~m}^{2}=50$
$\Rightarrow \mathrm{m}= \pm \dfrac{5}{2}$
S3: Let mid point be $(\mathrm{h}, \mathrm{k})$
Equation of chord whose mid point is given is
$\mathrm{T}=\mathrm{S} _{1}$
$\dfrac{\mathrm{xh}}{25}-\dfrac{\mathrm{yk}}{9}-1=\dfrac{\mathrm{b}^{2}}{25}-\dfrac{\mathrm{k}^{2}}{9}-1$
This passes through $(5,0) \dfrac{5 \mathrm{~h}}{25}-0-1=\dfrac{\mathrm{h}^{2}}{25}-\dfrac{\mathrm{K}^{2}}{9}-1$
$45 \mathrm{~h}=9 \mathrm{~h}^{2}-25 \mathrm{k}^{2}$
Locus of (h,k) is $9 x^{2}-45 x-25 y^{2}=0$
$\begin{aligned} & 9\left(x^{2}-5 x+\dfrac{25}{4}-\dfrac{25}{4}\right)-25 y^{2}=0 \\ & 9(x-5 / 2)^{2}-25 y^{2}=\dfrac{225}{4} \end{aligned}$
$\dfrac{(x-5 / 2)^{2}}{\left(\dfrac{25}{4}\right)}-\dfrac{y^{2}}{\left(\dfrac{9}{4}\right)}=1$
Length of latus rectum in not same.
S4: Since $(\operatorname{asec} \theta, b \tan \theta)$ lies on $y=m x+\sqrt{a^{2} m^{2}-b^{2}}$
$\therefore \operatorname{btan} \theta=\mathrm{amsec} \theta+\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}-\mathrm{b}^{2}}$
$(b \tan \theta-a \sec \theta)^{2}=a^{2} m^{2}-b^{2}$
$\mathrm{a}^{2} \mathrm{~m}^{2} \sin ^{2} \theta-$ eabmsin $\theta+\mathrm{b}^{2}=0 \quad(\because \cos \theta \neq 0)$
$(\mathrm{am} \sin \theta-\mathrm{b})^{2}=0$
$\therefore \sin \theta=\dfrac{\mathrm{b}}{\mathrm{am}}$
Hence $\theta=\sin ^{-1}\left(\dfrac{\mathrm{b}}{\mathrm{am}}\right)$
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 $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$ makes with its asymptotes is | (p) | 12 |
| (b) | If the line $y=3 x+\lambda$ touches the curve $9 x^{2}-5 y^{2}=45$, then $|\lambda|$ is | (q) | 6 |
| (c) | If the chord $\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}$ of the hyperbola $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{18}=1$ 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 $\lambda$ be the length of the latus rectum of the hyperbola $16 x^{2}-9 y^{2}+32 x+36 y-164=0$, then $3 \lambda$ is equal to | (s) | 32 |
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Solution
(a) Equation of tangent at $(a, 0)$ is $x=a$
asymptotes $y=\dfrac{b}{a} x$ & $y=-\dfrac{b}{a} x$
$\text { Area }=2 \times \dfrac{1}{2} \times \mathrm{a} \times \mathrm{b}=4 \times 3=12 {a}$
(b) $y=3 x+\lambda$ touches the curve $9 x^{2}-5 y^{2}=45$
$\begin{aligned} & \text { then } 9 x^2-5(3 x+\lambda)^2=45 \\ & \quad 36 x^2+30 \lambda x+5 \lambda^2+45=0 \text { has equal roots } \\ & \therefore 900 \lambda^2-4 \times 36\left(5 \lambda^2+45\right)=0 \\ & \lambda^2=36 \Rightarrow \lambda= \pm 6 \\ & \text { Hence }|\lambda|=6 \end{aligned}$
(c) The combined equation $\mathrm{OA}$ and $\mathrm{OB}$ we get when we make a homogeneous equation with the help of line and hyperbola
i.e. $18 \mathrm{x}^{2}-16 \mathrm{y}^{2}-288\left(\dfrac{\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha}{\mathrm{p}}\right)^{2}=0$
$9 p^{2} x^{2}-9 p^{2} y^{2}-144\left(x^{2} \cos ^{2} \alpha+y^{2} \sin ^{2} \alpha+2 x y \sin \alpha \cos \alpha\right)=0$
$\left(9 p^{2}-144 \cos ^{2} \alpha\right) x^{2}-288 \sin \alpha \cos \alpha \cdot x y+\left(-144 \sin ^{2} \alpha-8 p^{2}\right) y^{2}=0$
Lines are perpendicular $\Rightarrow 9 p^{2}-144 \cos ^{2} \alpha-8 p^{2}-144 \sin ^{2} \alpha=0$
$\begin{aligned} & \mathrm{p}^{2}=144 \\ & \mathrm{p}= \pm 12 \end{aligned}$
Radius of circle $=12$
$\therefore$ Diameter of circle $=24$
(c) $\rightarrow(\mathrm{r})$
(d) $16 x^{2}+32 x-9 y^{2}+36 y=164$
$16(\mathrm{x}+1)^{2}-9(\mathrm{y}-2)^{2}=144$
$\dfrac{(\mathrm{x}+1)^{2}}{9}-\dfrac{(\mathrm{y}-2)^{2}}{16}=1$
length of latus rectum $=2 \times \dfrac{16}{3}=\dfrac{32}{3}$
$\therefore 3 \lambda=32$
$(d) \rightarrow (s)$
Exercis - 9
1. For a given non zero value of $m$ each of the lines $\dfrac{x}{a}-\dfrac{y}{b}=m$ and $\dfrac{x}{a}+\dfrac{y}{b}=m$ meets the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ at a point. Sum of the ordinates of these points, is
(a) $\dfrac{a\left(1+m^{2}\right)}{m}$
(b) $\dfrac{a+b}{2 m}$
(c) $\dfrac{\mathrm{b}\left(1-\mathrm{m}^{2}\right)}{\mathrm{m}}$
(d) 0
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Answer: d2. For which of the hyperbola, we can have more than one pair of perpendicular tangents?
(a) $\dfrac{x^{2}}{4}-\dfrac{y^{2}}{9}=-1$
(b) $\dfrac{x^{2}}{4}-\dfrac{y^{2}}{9}=1$
(c) $x^{2}-y^{2}=9$
(d) $x y=9$
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Answer: a3. From point $(2,2)$ tangents are drawn to the hyperbola $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$ then point of contact lie in
(a) I & IV quadrants
(b) II & III quadrants
(c) III & IV quadrants
(d) II & III quadrants
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Answer: c4. If $(5,12)$ and $(24,7)$ are the foci of a conic passing through the origin then the eccentricity of conic is
(a) $\dfrac{\sqrt{386}}{13}$
(b) $\dfrac{\sqrt{386}}{12}$
(c) $\dfrac{\sqrt{386}}{38}$
(d) $\dfrac{\sqrt{386}}{25}$
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Answer: c5. For the hyperbola $9 x^{2}-16 y^{2}-18 x+32 y-151=0$
(a) directrix is $x=\dfrac{21}{5}$
(b) latus rectum $=\dfrac{9}{2}$
(c) eccentricity is $\dfrac{5}{4}$
(d) foci are $(6,1)$ and $(-4,1)$
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Answer: d6. 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 $\left(6, \dfrac{2}{3}\right)$ then coordinates of the fourth point are $\left(\dfrac{1}{4}, 16\right)$.
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
(c) Statement - 1 is True, Statement - 2 is False.
(d) Statement - 1 is False, statement 2 is True.
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Answer: dTrue/ 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 $\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$, 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
(c) TTFT
(d) TTTT
Comprehension
For the hyperbola $\dfrac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\dfrac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ the normal at $\mathrm{P}$ meets the transverse axis $\mathrm{A}^{\prime}$ in $\mathrm{D}$ and the conjugate axis $\mathrm{BB}^{\prime}$ in $\mathrm{E}$ and $\mathrm{CF}$ be perpendicular to the normal from the centre.
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Answer: b8. $\mathrm{PF} \times \mathrm{PD}=\mathrm{k}(\mathrm{CB})^{2}$, then $\mathrm{k}=$
(a) $\dfrac{1}{2}$
(b) 1
(c) 2
(d) 3
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Answer: b9. $\quad \mathrm{PF} \times \mathrm{PE}$ equal to
A program to give wings to girl students
(a) $\mathrm{CA} \times \mathrm{CB}$
(b) $(\mathrm{CB})^{2}$
(c) $(\mathrm{CA})^{2}$
(d) $(\mathrm{CF})^{2}$
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Answer: c10. Locus of middle point of $\mathrm{D}$ and $\mathrm{E}$ is a hyperbola of eccentricity
(a) $\sqrt{\mathrm{e}^{2}-1}$
(b) $\dfrac{1}{\sqrt{\mathrm{e}^{2}-1}}$
(c) $\mathrm{e} \sqrt{\mathrm{e}^{2}-1}$
(d) $\dfrac{\mathrm{e}}{\sqrt{\mathrm{e}^{2}-1}}$
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