Let a and b be non-zero and real numbers. Then, the equation $ (ax^2 + by^2 + c) \, ( x^2 - 5xy + 6y^2) = 0 $ represents
- four straight lines, when c = 0 and a, b are of the same sign
- two straight lines and a circle, when a = b and c is of sign opposite to that of a
- two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a
- a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a
The Correct Option is B
Solution and Explanation
Therefore the given equation
$ (ax^2 + by^2 + c) \, ( x^2 - 5xy + 6y^2) = 0 $ implies either
$ x^2 - 5 x y + 6 y^2 = 0 $
$\Rightarrow (x - 2y) \, (x - 3y) = 0 $
$\Rightarrow x = 2y $
and x = 3y
represent two straight lines passing through origin or
$ ax^2 + by^2 + c = 0 $ when c = 0 and a and b are of same
signs, then
$ ax^2 + by^2 + c = 0 $
x = 0
and y =- 0
which is a point specified as the origin.
When, a = b and c is of sign opposite to that of a
$ ax^2 + by^2 + c = 0 $ represents a circle.
Hence, the given equation
$\hspace22mm$ $ (ax^2 + by^2 + c) \, (x^2 - 5xy + 6 y^2 ) = 0 $
may represent two straight lines and a circle.
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c