HYPERBOLA- 7 (Rectangular Hyperbola)

Rectangular Hyperbola

A hyperbola whose asymptotes include a right angle is said to be rectangular hyperbola.

OR

If the lengths of transverse and conjugate axes of any hyperbola be equal it is called rectangular or equilateral hyperbola.

According to first definition

$ \begin{aligned} & 2 \tan ^{-1}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)=\frac{\pi}{2} \\ & \tan ^{-1} \frac{\mathrm{b}}{\mathrm{a}}=\frac{\pi}{4} \\ & \frac{\mathrm{b}}{\mathrm{a}}=\tan \frac{\pi}{4} \\ & \mathrm{~b}=\mathrm{a} \end{aligned} $

then $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ becomes $x^{2}-y^{2}=a^{2}$

According to second definition

When $\mathrm{a}=\mathrm{b}, \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ becomes $\mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{a}^{2}$

Eccentricity $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{2}$

Then asymptotes of $x^{2}-y^{2}=a^{2}$ are $x+y=0$ and $x-y=0$. Each of these two asymptotes is inclined at an angle of $45^{\circ}$ with the transverse axis. So, if we rotate the coordinate axes through an angle of $\frac{\pi}{4}$ keeping the origin fixed, then the axes coincide with the asymptotes of the hyperbola.

Now equation of asymptotes of new hyperbola is $x=0$

Then equation of hyperbola is $\mathrm{xy}=\mathrm{k}$ (constant)

The hyperbola passes through the point $\left(\frac{\mathrm{a}}{\sqrt{2}}, \frac{\mathrm{a}}{\sqrt{2}}\right)$

$ \therefore \mathrm{k}=\frac{\mathrm{a}^{2}}{2} $

Then equation of hyperbola is $x y=\frac{a^{2}}{2}$ or $x y=c^{2}$ where $c^{2}=\frac{a^{2}}{2}$

If the asymptotes of a rectangular hyperbola are $x=a, y=b$, then its equation is $(x-a)(y-b)=c^{2}$ $\mathbf{x y}=\mathbf{c}^{2}$

1. Asymptotes : $\mathrm{x}=0, \mathrm{y}=0$

2. Transverse axis: $y=x$

Conjugate axis : $\mathrm{y}=-\mathrm{x}$

3. Vertices $\mathrm{A}(\mathrm{c}, \mathrm{c}), \mathrm{A}^{\prime}(-\mathrm{c},-\mathrm{c})$

4. Foci : $\mathrm{S}(\mathrm{c} \sqrt{2}, \mathrm{c} \sqrt{2}), \mathrm{S}^{\prime}(-\mathrm{c} \sqrt{2},-\mathrm{c} \sqrt{2})$

5. Length of transverse axis $=\mathrm{AA}^{\prime}=2 \sqrt{2} \mathrm{c}$

6. Equation of auxiliary circle $x^{2}+y^{2}=2 c^{2}$

7. Equation of director circle $x^{2}+y^{2}=0$

8. $x^{2}-y^{2}=a^{2}$ and $x y=c^{2}$ intersect at right angles

Properties of Rectangular Hyperbola

1. Eccentricity of rectangular hyperbola is $\sqrt{2}$.

2. Since $x=c t, y=\frac{c}{t}$ satisfies $x y=c^{2}$

$\therefore(\mathrm{x}, \mathrm{y})=\left(\mathrm{ct}, \frac{\mathrm{c}}{\mathrm{t}}\right)(\mathrm{t} \neq 0)$ is called a ’ $\mathrm{t}$ ’ point on the rectangular hyperbola. The $\mathrm{x}=\mathrm{ct}, \mathrm{y}=\frac{\mathrm{c}}{\mathrm{t}}$ represents its parametric equation with parameter ’ $t$ '

3. Equation of chord joining $\mathrm{P}\left(\mathrm{Ct} _{1}, \frac{\mathrm{c}}{\mathrm{t} _{1}}\right)$ and $\mathrm{Q}\left(\mathrm{Ct} _{2}, \frac{\mathrm{c}}{\mathrm{t} _{2}}\right)$ is

$ \begin{aligned} & \mathrm{x}+\mathrm{yt} _{1} \mathrm{t} _{2}-\mathrm{c}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right)=0 \\ & \text { Slope of chord }=-\frac{1}{\mathrm{t} _{1} \mathrm{t} _{2}} \end{aligned} $

4. Equation of tangent at $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ is $\mathrm{xy} _{1}+\mathrm{yx} _{1}=2 \mathrm{c}^{2}$

5. Equation of tangent at is $\frac{x}{t}+y t=2 c$

$ \text { Slope of tangent }=-\frac{1}{t^{2}} $

6. Equation of normal at $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ is $\mathrm{xx} _{1}-\mathrm{yy} _{1}=\mathrm{x} _{1}{ }^{2}-\mathrm{y} _{1}{ }^{2}$

Equation of normal at is $\mathrm{xt}^{3}-\mathrm{yt}^{2}-\mathrm{ct}^{4}+\mathrm{c}=0$

$ \text { Slope of normal }=\mathrm{t}^{2} $

7. Point of intersection of tangents at $t _{1}$ and $t _{2}$ is

$ \left(\frac{2 \mathrm{ct} _{1} \mathrm{t} _{2}}{\mathrm{t} _{1}+\mathrm{t} _{2}}, \frac{2 \mathrm{c}}{\mathrm{t} _{1}+\mathrm{t} _{2}}\right) $

8. Point of intersection of normal at $t _{1}$ and $t _{2}$ is

$ \left(\frac{\mathrm{ct} _{1} \mathrm{t} _{2}\left(\mathrm{t} _{1}{ }^{2}+\mathrm{t} _{1} \mathrm{t} _{2}+\mathrm{t} _{2}^{2}\right)-\mathrm{c}}{\mathrm{t} _{1} \cdot \mathrm{t} _{2}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right)}, \frac{\mathrm{ct} _{1}{ }^{3} \mathrm{t} _{2}{ }^{3}+\mathrm{c}\left(\mathrm{t} _{1}{ }^{2}+\mathrm{t} _{1} \mathrm{t} _{2}+\mathrm{t} _{2}{ }^{2}\right)}{\mathrm{t} _{1} \cdot \mathrm{t} _{2}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right)}\right) $

Examples

1. If the normal at the point $\mathrm{t} _{1}$ to the rectangular hyperbola $\mathrm{xy}=\mathrm{c}^{2}$ meets it again at the point $\mathrm{t} _{2}$ then

(a) $\mathrm{t} _{1} \mathrm{t} _{2}=-1$

(b) $\mathrm{t} _{1}^{3 \mathrm{t} _{2}}=-1$

(c) $\mathrm{t} _{1} \mathrm{t} _{2}{ }^{3}=-1$

(d) $\mathrm{t} _{1}{ }^{2} \mathrm{t} _{2}{ }^{2}=-1$

Show Answer

Solution:

Equation of normal at $\left(\mathrm{ct} _{1}, \frac{\mathrm{c}}{\mathrm{t} _{1}}\right)$ to the hyperbola $x y=c^{2}$ is

$ \mathrm{xt} _{1}^{3}-\mathrm{yt} _{1}-\mathrm{ct} _{1}^{4}+\mathrm{c}=0 $

But this passes through $\left(\mathrm{ct} _{2}, \frac{\mathrm{c}}{\mathrm{t} _{2}}\right)$ then

$ \begin{aligned} & \mathrm{ct} _{2} \mathrm{t} _{1}^{3}-\frac{\mathrm{c}}{\mathrm{t} _{2}} \mathrm{t} _{1}-\mathrm{ct} _{1}{ }^{4}+\mathrm{c}=0 \\ & \mathrm{t} _{2} \mathrm{t} _{1}{ }^{3}-\frac{\mathrm{t} _{1}}{\mathrm{t} _{2}}-\mathrm{t} _{1}{ }^{4}+1=0 \\ & \mathrm{t} _{2}{ }^{2} \mathrm{t} _{1}{ }^{3}-\mathrm{t} _{1}-\mathrm{t} _{1}{ }^{4} \mathrm{t} _{2}+\mathrm{t} _{2}=0 \\ & \mathrm{t} _{1} \mathrm{t} _{2}\left(\mathrm{t} _{2}-\mathrm{t} _{1}\right)+1\left(\mathrm{t} _{2}-\mathrm{t} _{1}\right)=0 \\ & \left(\mathrm{t} _{2}-\mathrm{t} _{1}\right)\left(\mathrm{t} _{1}{ }^{3} \mathrm{t} _{2}+1\right)=0 \\ & \Rightarrow \mathrm{t} _{1} \mathrm{t} _{2}=-1 \quad\left(\because \mathrm{t} _{1} \neq \mathrm{t} _{2}\right) \end{aligned} $

Answer: (b)

2. A triangle has its vertices on a rectangular hyperbola $\mathrm{xy}=\mathrm{c}^{2}$

The orthocenter of the triangle lies on

(a) $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}$

(b) $x^{2}-y^{2}=c^{2}$

(c) $x y=c^{2}$

(d) None of these

Show Answer

Solution:

Let $\mathrm{A}\left(\mathrm{ct} _{1}, \frac{\mathrm{c}}{\mathrm{t} _{1}}\right) \mathrm{B}\left(\mathrm{ct} _{2}, \frac{\mathrm{c}}{\mathrm{t} _{2}}\right) \mathrm{C}\left(\mathrm{ct} _{3}, \frac{\mathrm{c}}{\mathrm{t} _{3}}\right)$ be the vertices of atriangle lies on $\mathrm{xy}=\mathrm{c}^{2}$

Now slope of $\mathrm{BC}=\frac{\left(\frac{\mathrm{c}}{\mathrm{t} _{3}}-\frac{\mathrm{c}}{\mathrm{t} _{2}}\right)}{\left(\mathrm{ct} _{3}-\mathrm{ct} _{2}\right)}=-\frac{1}{\mathrm{t} _{2} \mathrm{t} _{3}}$

Hence slope of $A D=t _{2} t _{3}$

Equation of $\mathrm{AD}$ is

$\begin{aligned} & \mathrm{y}-\frac{\mathrm{c}}{\mathrm{t}_1}=\mathrm{t}_2 \mathrm{t}_3\left(\mathrm{x}-\mathrm{ct}_1\right) \\ & \mathrm{t}_1 \mathrm{y}-\mathrm{c}=\mathrm{xt}_1 \mathrm{t}_2 \mathrm{t}_3-\mathrm{ct}_1^2 \mathrm{t}_2 \mathrm{t}_3……….(1)\end{aligned}$

Similarly

$ \text { slope of } A C=-\frac{1}{t _{1} t _{3}} $

slope of $\mathrm{BE}=\mathrm{t} _{1} \mathrm{t} _{3}$

Equation of $\mathrm{BE}$ is $\mathrm{t} _{2} \mathrm{y}-\mathrm{c}=\mathrm{xt} _{1} \mathrm{t} _{2} \mathrm{t} _{3}-\mathrm{ct} _{2}{ }^{2} \mathrm{t} _{1} \mathrm{t} _{3}……..(2)$

Solving (1) and (2) we get $\mathrm{y}=-\mathrm{ct} _{1} \mathrm{t} _{2} \mathrm{t} _{3} ~ \& ~ \mathrm{x}=-\frac{\mathrm{c}}{\mathrm{t} _{1} \mathrm{t} _{2} \mathrm{t} _{3}}$

Which satisfies $\mathrm{xy}=\mathrm{c}^{2}$

Therefore orthocentre lies on $\mathrm{xy}=\mathrm{c}^{2}$

Answer: (c)

3. If $\mathrm{PN}$ is the perpendicular from a point on a rectangular hyperbola $\mathrm{xy}=\mathrm{c}^{2}$ to its asymptotes then locus of the midpoint of $\mathrm{PN}$ is

(a) $x y=\frac{c^{2}}{2}$

(b) $x y=\frac{c^{2}}{4}$

(c) $x y=2 c^{2}$

Show Answer

Solution:

Let $P\left(x _{1}, y _{1}\right)$ be a point on $x y=c^{2}$ Le $Q(h, k)$ be the Midpoint of $P N$ then $x _{1}=h$ and $k=\frac{y _{1}}{2}$

$(\mathrm{x}, \mathrm{y}) _{1}$ lies on $\mathrm{xy}=\mathrm{c}^{2} \Rightarrow \mathrm{h}(2 \mathrm{k})=\mathrm{c}^{2}$ Locus of

$(h, k)$ is $2 x y=c^{2}$

Answer: (a)

4. $\mathrm{PQ}$ and RS are two perpendicular chords of the rectangular hyperbola $\mathrm{xy}=\mathrm{c}^{2}$. IfO is the centre of the rectangular hyperbola, then product of the slopes of $\mathrm{OP}, \mathrm{OQ}, \mathrm{OR}$ and $\mathrm{OS}$ is

(a) $-1$

(b) $-2$

(c) $1$

(d) $2$

Show Answer

Solution:

Let coordinates of P, Q, R, S be $\left(\mathrm{ct} _{\mathrm{i}}, \frac{\mathrm{c}}{\mathrm{t} _{\mathrm{i}}}\right), \mathrm{i}=1,2,3,4$

Now PQ $\perp$ RS

$ \Rightarrow \mathrm{m} _{\mathrm{PQ}} \times \mathrm{m} _{\mathrm{RS}}=-1 $

$ \Rightarrow \frac{\frac{\mathrm{c}}{\mathrm{t} _{2}}-\frac{\mathrm{c}}{\mathrm{t} _{1}}}{\mathrm{ct} _{2}-\mathrm{ct} _{1}} \times \frac{\frac{\mathrm{c}}{\mathrm{t} _{4}}-\frac{\mathrm{c}}{\mathrm{t} _{3}}}{\mathrm{ct} _{4}-\mathrm{ct} _{3}}=-1 $

$ \begin{aligned} & \Rightarrow-\frac{1}{\mathrm{t} _{1} \mathrm{t} _{2}} \times \frac{-1}{\mathrm{t} _{3} \mathrm{t} _{4}}=-1 \\ & \Rightarrow \mathrm{t} _{1} \mathrm{t} _{2} \mathrm{t} _{3} \mathrm{t} _{4}=-1 \end{aligned} $

Now slope of $\mathrm{OP}=\frac{\frac{\mathrm{c}}{\mathrm{t} _{1}}}{\mathrm{ct} _{1}}=\frac{1}{\mathrm{t} _{1}{ }^{2}}$

Similarly slope of $\mathrm{OQ}=\frac{1}{\mathrm{t} _{2}{ }^{2}}$

Similarly slope of $\mathrm{OR}=\frac{1}{\mathrm{t} _{3}{ }^{2}}$

Similarly slope of $O S=\frac{1}{\mathrm{t} _{4}{ }^{2}}$

Product of their slopes $=\frac{1}{\mathrm{t} _{1}{ }^{2} \cdot \mathrm{t} _{2}{ }^{2} \cdot \mathrm{t} _{3}{ }^{2} \cdot \mathrm{t} _{4}{ }^{2}}=\frac{1}{(-1)^{2}}=1$

Answer: (c)

5. The angle between the rectangular hyperbolas

$(y-m x)(m y+x)=a^{2}$ and $\left(m^{2}-1\right)\left(y^{2}-x^{2}\right)+4 m x y=b^{2}$ is

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

(b) $\frac{\pi}{3}$

(c) $\frac{\pi}{4}$

(d) $\tan ^{-1}\left(\frac{b}{a}\right)$

Show Answer

Solution:

$\begin{aligned} & (y-m x)(m y+x)=a^2 \\ & \left(\frac{d y}{d x}-m\right)(m y+x)+\left(m \frac{d y}{d x}+1\right)(y-m x)=0 \\ & (m y+x) \frac{d y}{d x}+m(y-m x) \frac{d y}{d x}=m(m y+x)-(y-m x) \\ & \frac{d y}{d x}=\frac{m^2 y+m x-y+m x}{m y+x+m y-m^2 x}=\frac{m^2 y+2 m x-y}{-m^2 x+x+2 m y}=m_{1-}…………(1)\end{aligned}$

For another hyperbola

$\begin{aligned} & \left(m^2-1\right)\left(y^2-x^2\right)+4 m x y=b^2 \\ & \left(m^2-1\right)\left(2 y \frac{d y}{d x}-2 x\right)+4 m\left(y+x \frac{d y}{d x}\right)=0 \\ & \left(m^2-1\right) y \frac{d y}{d x}-x\left(m^2-1\right)+2 m y+2 m x \frac{d y}{d x}=0 \\ & \left(m^2 y-y+2 m x\right) \frac{d y}{d x}=m^2 x-x-2 m y \\ & \frac{d y}{d x}=\frac{m^2 x-x-2 m y}{m^2 y+2 m x-y}=m_2……….(2)\end{aligned}$

Now $\mathrm{m} _{1} \times \mathrm{m} _{2}=-1$

$ \therefore \text { angle }=\frac{\pi}{2} $

Answer: (a)

6. The family of the curves which intersect the family of rectangular hyperbola $x y=c^{2}$ orthogonally is

(a) family of circle

(b) family of parabola

(c) family of ellipse

(d) family of hyperbola

Show Answer

Solution:

$\mathrm{xy}=\mathrm{c}^{2}$

Differentiate w.r.t. $\mathrm{x}$

$ \mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=0 $

Replace $\frac{d y}{d x}$ by $-\frac{d x}{d y}$, we get

$ \begin{aligned} & y-x \frac{d x}{d y}=0 \\ & y d y-x d x=0 \\ & y^{2}-x^{2}=k \text { (where } k \text { is paramenter) } \end{aligned} $

family of hyperbola.

Answer: (d)

Practice questions

1. The coordinates of the foci of the rectangular hyperbola $x y=c^{2}$ is

(a) $( \pm \mathrm{c}, \pm \mathrm{c})$

(b) $\left( \pm \frac{\mathrm{c}}{\sqrt{2}}, \pm \frac{\mathrm{c}}{\sqrt{2}}\right)$

(c) $( \pm \mathrm{c} \sqrt{2}, \pm \mathrm{c} \sqrt{2})$

(d) $\left( \pm \frac{\mathrm{c}}{2}, \pm \frac{\mathrm{c}}{2}\right)$

Show Answer Answer: (c)

2. The equation of directories of the rectangular hyperbola $x y=c^{2}$ is

(a) $x+y= \pm \mathrm{c}$

(b) $x+y= \pm c \sqrt{c}$

(c) $x+y= \pm \frac{c}{\sqrt{c}}$

(d) $x \pm y=0$

Show Answer Answer: (b)

3. If the line $a x+b y+c=0$ is a normal to the hyperbola $x y=1$, then

(a) $a>0, b>0$

(b) $\mathrm{a}<0, \mathrm{~b}<0$

(c) $a>0, b<0$

(d) $\mathrm{a}<\mathrm{o}, \mathrm{b}>0$

Show Answer Answer: (c, d)

4. Consider the set of hyperbolas $x y=k, k \in R$. Let $e _{1}$, be the eccentricity when $k=4$ and $e _{2}$ be the eccentricity when $\mathrm{k}=9$, then $\mathrm{e} _{1}-\mathrm{e} _{2}$ is equal to

(a) $0$

(b) $1$

(c) $\frac{3}{2}$

(d) $2$

Show Answer Answer: (a)

5. If chords of the hyperbola $x^{2}-y^{2}=a^{2}$ touch the parabola $y^{2}=4 a x$ then the locus of the middle points of these chords in the crane

(a) $\mathrm{y}^{2}=\mathrm{x}^{3}$

(b) $y^{2}(x-a)=x^{2}$

(c) $y^{3}(x-a)=x^{2}$

(d) $y^{2}(x-a)=x^{3}$

Show Answer Answer: (d)

6. If the tangent and normal to a rectangular hyperbola $x y=4$ cut off intercepts $a _{1}$ and $a _{2}$ on one axis and $b _{1}$ and $b _{2}$ on the other, then $a _{1} a _{2}+b _{1} b _{2}$ is

(a) $-4$

(b) $0$

(c) $1$

(d) $4$

Show Answer Answer: (b)

7. The points of intersection of the cranes whose parametric equations are $x=t^{2}+1, y=2 t$ and $x=25$, $\mathrm{y}=\frac{2}{5}$ is given by

(a) $(1, 2)$

(b) $(2, 2)$

(c) $(1, -3)$

(d) $(-2, 4)$

Show Answer Answer: (b)

8. Number of maximum tangents from any point to the hyperbola $\mathrm{xy}=\mathrm{c}^{2}$ is

(a) $1$

(b) $2$

(c) $3$

(d) $4$

Show Answer Answer: (b)

9. The length of the latus rectum of the hyperbola $x y-3 x-3 y+7=0$ is

(a) $2$

(b) $4$

(c) $2 \sqrt{2}$

(d) None of these

Show Answer Answer: (b)

10. If $x=9$ is the chord of contact of the hyperbola $x^{2}-y^{2}=9$, then the equation of the corresponding pair of tangent is

(a) $9 \mathrm{x}^{2}-8 \mathrm{y}^{2}-18 \mathrm{x}+9=0$

(b) $9 x^{2}-8 y^{2}-18 x-9=0$

(c) $9 x^{2}-8 y^{2}+18 x+9=0$

(d) $9 x^{2}-8 y^{2}+18 x-9=0$

Show Answer Answer: (a)

11. The equation of the chord joining two points $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ and $\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right)$ on the rectangular hyperbola $\mathrm{xy}=$ $\mathrm{c}^{2}$ is

(a) $\frac{x}{x _{1}-x _{2}}+\frac{y}{y _{1}-y _{2}}=1$

(b) $\frac{x}{y _{1}+y _{2}}+\frac{y}{x _{1}+x _{2}}=1$

(c) $\frac{x}{x _{1}+x _{2}}+\frac{y}{y _{1}+y _{2}}=1$

(d) $\frac{x}{y _{1}-y _{2}}+\frac{y}{x _{1}-x _{2}}=1$

Show Answer Answer: (c)


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