Question
Question: The equation \[{y^2} - {x^2} + 2x - 1 = 0\] represents ? A. A pair of straight lines B. A Circl...
The equation y2−x2+2x−1=0 represents ?
A. A pair of straight lines
B. A Circle
C. A Parabola
D. Ellipse
Solution
For the questions related to the conic section always compare the given equation with the general equation of conic section which is ax2+ 2hxy + by2+ 2gx + 2fy + c = 0 if the specified conic section ( circle , parabola ) etc. is not given .
Complete step by step answer:
Given : y2−x2+2x−1=0
On comparing the given equation with the general equation ax2+ 2hxy + by2+ 2gx + 2fy + c = 0 , we get
a=−1 , b=1 , g=1 , h=0 , f=0 , c=−1 . Now , for different conic sections we have different conditions . For that we have to find the determinant ( Δ ) of the general equation then we will have conditions .
Determinant of the general equation will be \Delta = \left( {\begin{array}{*{20}{c}}
{{a_{}}}&h;&g; \\\
h&b;&f; \\\
g&f;&c;
\end{array}} \right) , if Δ the will become 0 . Then , it is a pair of straight lines . Otherwise , if Δ is not 0 ( Δ=0 ) . Then , we have
If h2=ab , then it is a parabola .
If h2>ab , then it is a hyperbola .
If h2=0,a=b , then it is a circle .
If h2>ab , then it is an ellipse .
Putting the values of the general equations co – efficients , we get \Delta = \left( {\begin{array}{*{20}{c}}
{ - {1_{}}}&0&1 \\\
0&1&0 \\\
1&0&{ - 1}
\end{array}} \right) , on solving the determinant , we get
= −1(−1−0)−0+1(0−1)
= −1(−1)−0+1(−1) , on solving we get
= 1−1
= 0 .
Therefore , we get the Δ as Δ=0. So , the given equation represents a pair of straight lines .
So, the correct answer is “Option A”.
Note: The general equation for the conic sections for the is a second degree homogeneous equation where a , h , b does not vary simultaneously . The determinant ( Δ ) from the general equation is obtained using the equation abc + 2fgh − af2− af2− bg2− ch2= 0.