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 vertex, focus, end points of the latus rectum, the length of latus rectum, the equation of axis and directrix of the parabolas $2 x^{2} - 4 y + 6 x - 3 = 0$

Accepted Answer

We have, $2 x^{2} - 4 y + 6 x - 3 = 0$

$\Rightarrow 2 (x^{2} + 3 x) = 4 y + 3$

$\Rightarrow 2 (x^{2} + 3 x + \frac{9}{4}) = 4 y + 3 + \frac{18}{4}$

$\Rightarrow {(x + \frac{3}{2})}^{2} = 2 y + \frac{15}{4}$

$\Rightarrow {(x + \frac{3}{2})}^{2} = 2 (y + \frac{15}{8})$ $\dots (1)$

Shifting the origin to the point $(- \frac{3}{2}, - \frac{15}{8})$ without rotating the axes and denoting the new coordinates with respect to these new axes by $X$ and $Y$ , we have

$x = X - \frac{3}{2}$ and $y = Y - \frac{15}{8}$ $\dots (2)$

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

$X^{2} = 2 Y$ $\dots (3)$

This is of the form $X^{2} = 4 a Y$ . On comparing, we get $4 a = 2 \Rightarrow a = \frac{1}{2}$ .

Vertex: The coordinates of the vertex with respect to the new axes are $(X = 0, Y = 0)$ .

So, the coordinates of the vertex with respect to the old axes are $(- \frac{3}{2}, - \frac{15}{8})$ .

Axis: The equation of the parabola $(3)$ with respect to the new axes is $X = 0$ .

So, the equation of the axis with respect to the old axis is $x = - \frac{3}{2}$ $\Rightarrow 2 x + 3 = 0$ [Putting $X = 0$ in $(2)$ ]

Focus: The coordinates of the focus with respect to the new axes are $(X = 0, Y = a)$ i.e, $(X = 0, Y = \frac{1}{2})$ .

So, the coordinates of the focus with respect to the old axes are $(- \frac{3}{2}, - \frac{11}{8})$ [Putting $X = 0, Y = \frac{1}{2}$ in $(2)$ ]

Directrix: The equation of the directrix with respect to the new axes is $Y = - a$ i.e., $Y = - \frac{1}{2}$ .

So, the equation of the directrix with respect to the old axes is $y = - \frac{19}{8}$ $\Rightarrow 8 y + 19 = 0$ [Putting $Y = - \frac{1}{2}$ in $(2)$ ]

Latus-rectum: The length of the latus-rectum of the parabola $(3)$ is equal to $4 a = 4 (\frac{1}{2}) = 2$

End-points of Latus rectum: End points of latus rectum with respect to the new axes are $(X = \pm 2 a, Y = a)$ i.e., $(X = \pm 1, Y = \frac{1}{2})$ .

So, the end points of latus rectum with respect to old axes are $(- \frac{1}{2}, - \frac{11}{8})$ and $(- \frac{5}{2}, - \frac{11}{8})$ [Putting $X = \pm 1, Y = \frac{1}{2}$ in $(2)$ ].

Reducing to standard form, obtain the co-ordinates of the vertex, focus, end points of the latus rectum, the length of latus rectum, the equation of axis and directrix of the parabolas $2 x^{2} - 4 y + 6 x - 3 = 0$

Important Questions on Conic Sections

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Reducing to standard form, obtain the co-ordinates of the vertex, focus, end points of the latus rectum, the length of latus rectum, the equation of axis and directrix of the parabolas 2x2-4y+6x-3=0

Important Questions on Conic Sections

Learn and Prepare for any exam you want !

Reducing to standard form, obtain the co-ordinates of the vertex, focus, end points of the latus rectum, the length of latus rectum, the equation of axis and directrix of the parabolas $2 x^{2} - 4 y + 6 x - 3 = 0$