Question:
The axis of the parabola $x^2 + 2xy + y^2 - 5x + 5y - 5 = 0$ is
The axis of the parabola $x^2 + 2xy + y^2 - 5x + 5y - 5 = 0$ is
Updated On: Apr 27, 2024
- $x + y = 0$
- $x + y - 1 = 0$
- $x - y + 1 = 0$
- $x-y=\frac{1}{\sqrt{2}}$
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The Correct Option is A
Solution and Explanation
Given, equation
$ x^{2}+2 x y+y^{2}-5 x+5 y-5=0$
$\Rightarrow (x+y)^{2}=5 x-5 y+5 $
$\Rightarrow (x+y)^{2}=5(x-y+1)$
$\therefore$ Axis of the parabola is $x+y=0$
$ x^{2}+2 x y+y^{2}-5 x+5 y-5=0$
$\Rightarrow (x+y)^{2}=5 x-5 y+5 $
$\Rightarrow (x+y)^{2}=5(x-y+1)$
$\therefore$ Axis of the parabola is $x+y=0$
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.