Name: Normal to a Circle
Uploaded: 2019-07-19
Duration: 2 min 45 s
Description: The normal to a circle is a straight line drawn at 90 degrees to the tangent at the point where the tangent touches the circle. Normal of a circle passes thr...

Question

Reducing to standard form, obtain the co-ordinates of the centre, the vertices, the foci, the end points of conjugate axis, the end points of latera recta, the equation of directrices and the eccentricity of the hyperbolas $9 y^{2} - 4 x^{2} - 90 y + 189 = 0$

Accepted Answer

We have, $9 y^{2} - 4 x^{2} - 90 y + 189 = 0$

$\Rightarrow 9 (y^{2} - 10 y) - 4 x^{2} = - 189$

$\Rightarrow 9 (y^{2} - 10 y + 25) - 4 {(x - 0)}^{2} = - 189 + 225$

$\Rightarrow 9 {(y - 5)}^{2} - 4 {(x - 0)}^{2} = 36$

$\Rightarrow \frac{{(y - 5)}^{2}}{4} - \frac{{(x - 0)}^{2}}{9} = 1$ $\dots (1)$

Shifting the origin at $(0, 5)$ without rotating the axes and denoting the new coordinates with respect to these axes by $Y$ and $X$ , we have

$x = X$ and $y = Y + 5$ $\dots (2)$

Using these relations, equation $(1)$ reduces to

$\frac{Y^{2}}{4} - \frac{X^{2}}{9} = 1$ $\dots (3)$

This is of the form $\frac{Y^{2}}{b^{2}} - \frac{X^{2}}{a^{2}} = 1$ , where $a = 3$ and $b = 2$ . So, we have

Eccentricity: The eccentricity $e$ is given by

$e = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}$

Vertices: The coordinates of the vertices with respect to the old axes are $(X = 0, Y = \pm b)$ i.e, $(X = 0, Y = \pm 2)$ .

So, the coordinates of the vertices with respect to the old axes are $(0, \pm 2 + 5)$ [Putting $X = 0, Y = \pm 2$ in $(2)$ ].

Foci: The coordinates of the foci with respect to the old axes are $(X = 0, Y = \pm b e)$ i.e., $(X = 0, Y = \pm \sqrt{13})$ .

So, the coordinates of the foci with respect to the old axes are $(0, \pm \sqrt{13} + 5)$ [Putting $X = 0, Y = \pm \sqrt{13}$ in $(2)$ ]

Directrices: The equation of the directrices with respect to the new axes are $Y = \pm \frac{b}{e}$ i.e., $Y = \pm \frac{2}{\frac{\sqrt{13}}{2}} = \pm \frac{4}{\sqrt{13}}$

So, the equation of the directrices with respect to the old axes are

$y = \pm \frac{4}{\sqrt{13}} + 5$

End-points of conjugate axis: End-points of conjugate axis with respect to new axes are $(X = \pm a, Y = 0)$ i.e, $(X = \pm 3, Y = 0)$

So, the end-points of conjugate axis with respect to the old axes are $(\pm 3, 5)$ .

End-points of Latus rectum: End-points of latus rectum with respect to new axes are $(X = \pm \frac{a^{2}}{b}, Y = \pm b e)$ i.e, $(X = \pm \frac{9}{2}, Y = \sqrt{13})$ .

So, the end-points of latus rectum with respect to the old axes are $(\pm \frac{9}{2}, \sqrt{13} + 5)$ .

Reducing to standard form, obtain the co-ordinates of the centre, the vertices, the foci, the end points of conjugate axis, the end points of latera recta, the equation of directrices and the eccentricity of the hyperbolas $9 y^{2} - 4 x^{2} - 90 y + 189 = 0$

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Reducing to standard form, obtain the co-ordinates of the centre, the vertices, the foci, the end points of conjugate axis, the end points of latera recta, the equation of directrices and the eccentricity of the hyperbolas 9y2-4x2-90y+189=0

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Reducing to standard form, obtain the co-ordinates of the centre, the vertices, the foci, the end points of conjugate axis, the end points of latera recta, the equation of directrices and the eccentricity of the hyperbolas $9 y^{2} - 4 x^{2} - 90 y + 189 = 0$