Question 15 Marks
Show that the equation $x^2-2 y^2-2 x+8 y-1=0$ represents a hyperbola. Find the coordinates of the centre, lengths of the axes, eccentricity, latusrectum, coordinates of foci and vertices and equations of directrices of the hyperbola.
Answer
View full question & answer→We have,
$x^2-2 y^2-2 x+8 y-1=0$
$\begin{array}{l}\Rightarrow\left(x^2-2 x\right)-2\left(y^2-4 y\right)=1 \\ \Rightarrow\left(x^2-2 x+1\right)-2\left(y^2-4 y+4\right)=-6 \\ \Rightarrow(1-x)^2-2(y-2)^2=-6 \\ \Rightarrow \frac{(x-1)^2}{(\sqrt{6})^2}-\frac{(y-2)^2}{(\sqrt{3})}=-1 \ldots \text { (i) }\end{array}$
Shifting the origin at (1, 2) without rotating the coordinate axes and denoting the new coordinates with respect to these axes by X
and Y, we obtain
x = X + 1 and y = Y + 2 ... (ii)
Using these relations, equation (i) reduces to
$\frac{X^2}{(\sqrt{6})^2}-\frac{Y^2}{(\sqrt{3})^2}=-1 \ldots$ (iii)
Comparing equation (iii) with standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$, we get
$\begin{array}{l}a^2=(\sqrt{6})^2 \text { and } b^2=(\sqrt{3})^2 \\ a=(\sqrt{6}) \text { and } b=(\sqrt{3})\end{array}$
Centre
The coordinates of the centre with respect to the new axes are (X = 0, Y = 0).
So, the coordinates of the centre with respect to the old axes are
$(1,2)[$ Putting $X =0, Y =0$ in (ii)]
Lengths of the axes:
$\therefore$ Transverse axis $=2 b=2 \sqrt{3}$ and, Conjugate axis $=2 a =2 \sqrt{6}$.
Eccentricity:
$e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{6}{3}}=\sqrt{3}$
Latusrectum:
Foci:
The coordinates of foci with respect to the new axes are ( $X=0, Y= \pm$ be) i.e. $(X=0, y= \pm 3$ ). So, the coordinates of foci with respect to the old axes are
$(1,2 \pm 3)$ i.e. $(1,5)$ and $(1,-1)[$ Putting $X=0, y= \pm 3$ in (ii)]
Directrices:
The coordinates of the vertices with respect to the new axes are $X=0, Y= \pm b)$ i.e. $(x=0, y= \pm \sqrt{3})$ So, the coordinates of the vertices with respect to the old axes are $(1,2 \pm \sqrt{3})$ i.e. $(1,2+\sqrt{3})$ and $(1,2-\sqrt{3})$ [Putting $X=0, Y= \pm \sqrt{3}$ in (ii)]
Directrices:
The equations of the directrices with respect to the new axes are $Y = \pm \frac{b}{e}$ i.e. $y = \pm 1$.
So, the equations of the directrices with respect to the old axes are $y=2 \pm 1$ i.e. $y=1$ and $y=3[$ Putting $Y= \pm 2$ in (ii) $]$
$x^2-2 y^2-2 x+8 y-1=0$
$\begin{array}{l}\Rightarrow\left(x^2-2 x\right)-2\left(y^2-4 y\right)=1 \\ \Rightarrow\left(x^2-2 x+1\right)-2\left(y^2-4 y+4\right)=-6 \\ \Rightarrow(1-x)^2-2(y-2)^2=-6 \\ \Rightarrow \frac{(x-1)^2}{(\sqrt{6})^2}-\frac{(y-2)^2}{(\sqrt{3})}=-1 \ldots \text { (i) }\end{array}$
Shifting the origin at (1, 2) without rotating the coordinate axes and denoting the new coordinates with respect to these axes by X
and Y, we obtain
x = X + 1 and y = Y + 2 ... (ii)
Using these relations, equation (i) reduces to
$\frac{X^2}{(\sqrt{6})^2}-\frac{Y^2}{(\sqrt{3})^2}=-1 \ldots$ (iii)
Comparing equation (iii) with standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$, we get
$\begin{array}{l}a^2=(\sqrt{6})^2 \text { and } b^2=(\sqrt{3})^2 \\ a=(\sqrt{6}) \text { and } b=(\sqrt{3})\end{array}$
Centre
The coordinates of the centre with respect to the new axes are (X = 0, Y = 0).
So, the coordinates of the centre with respect to the old axes are
$(1,2)[$ Putting $X =0, Y =0$ in (ii)]
Lengths of the axes:
$\therefore$ Transverse axis $=2 b=2 \sqrt{3}$ and, Conjugate axis $=2 a =2 \sqrt{6}$.
Eccentricity:
$e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{6}{3}}=\sqrt{3}$
Latusrectum:
Foci:
The coordinates of foci with respect to the new axes are ( $X=0, Y= \pm$ be) i.e. $(X=0, y= \pm 3$ ). So, the coordinates of foci with respect to the old axes are
$(1,2 \pm 3)$ i.e. $(1,5)$ and $(1,-1)[$ Putting $X=0, y= \pm 3$ in (ii)]
Directrices:
The coordinates of the vertices with respect to the new axes are $X=0, Y= \pm b)$ i.e. $(x=0, y= \pm \sqrt{3})$ So, the coordinates of the vertices with respect to the old axes are $(1,2 \pm \sqrt{3})$ i.e. $(1,2+\sqrt{3})$ and $(1,2-\sqrt{3})$ [Putting $X=0, Y= \pm \sqrt{3}$ in (ii)]
Directrices:
The equations of the directrices with respect to the new axes are $Y = \pm \frac{b}{e}$ i.e. $y = \pm 1$.
So, the equations of the directrices with respect to the old axes are $y=2 \pm 1$ i.e. $y=1$ and $y=3[$ Putting $Y= \pm 2$ in (ii) $]$
