Question

The image of the pair of lines represented by \[a{x^2} + 2hxy + b{y^2} = 0\]by the line mirror $y = 0$ is
$A)a{x^2} - 2hxy - b{y^2} = 0$
$B)b{x^2} - 2hxy - a{y^2} = 0$
$C)b{x^2} + 2hxy + a{y^2} = 0$
$D)a{x^2} - 2hxy + b{y^2} = 0$

Answer 1

Hint:First, from the given that we have an equation represented by \[a{x^2} + 2hxy + b{y^2} = 0\]and then using this equation we need to find the image of the pair of line equation from any one of the options is correct.
We use the mirror line as $y = 0$and we are not going to substitute this in the original equation, because the value of the $y$ is given as zero.
We put the value into some terms by using the addition and multiplication operation and then we substitute the value into the given equation to get the image of the equation.

Complete step-by-step solution:
Since given that we have the equation \[a{x^2} + 2hxy + b{y^2} = 0\]and we need to find its image using the information line mirror $y = 0$
By the use of multiplication property, multiply both sides by the number $2$on the given mirror line, then we get $y = 0 \Rightarrow 2y = 2 \times 0$
Since any number or value that multiplies into the number zero will get the resultant zero only, thus we get the mirror value as $y = 0 \Rightarrow 2y = 2 \times 0 \Rightarrow 2y = 0$
Now, by the use of the addition property that sum of the given two numbers or values yields a new value, and likewise a value can be separated into equal parts as $2y = y + y$
Hence, we get the mirror value as $y = 0 \Rightarrow 2y = 0 \Rightarrow y + y = 0$
Now, by the subtraction operation, subtract $ - y$on both sides, then we get $y + y = 0 \Rightarrow y + y - y = 0 - y$
Further solving this we get the mirror line as $y + y - y = 0 - y \Rightarrow y = - y$
Therefore, we get the mirror line as $y = - y$
Now substitute the value of the mirror line into the given original equation represented \[a{x^2} + 2hxy + b{y^2} = 0\]
Thus, we get \[a{x^2} + 2hx( - y) + b{( - y)^2} = 0\]
Since the square root of negative values get the positive like $( - y){}^2 = - y \times ( - y) = {y^2}$
Therefore, we get the equation as \[a{x^2} + 2hx( - y) + b{( - y)^2} = 0 \Rightarrow a{x^2} - 2hxy + b{y^2} = 0\]
Thus, the option $D)a{x^2} - 2hxy + b{y^2} = 0$is correct and is the image of the given equation.

Note:We can also able to solve the given problem by taking the image line as $x = - x$ from $x = 0$then proceed the same method we get, \[a{( - x)^2} + 2h( - x)y + b{y^2} = 0 \Rightarrow a{x^2} - 2hxy + b{y^2} = 0\]and hence we get the same result.
Since there is no possibility for the options $B)b{x^2} - 2hxy - a{y^2} = 0$$C)b{x^2} + 2hxy + a{y^2} = 0$because the image of the line only varies the variables on $x$ or $y$ not the constant $a,b$ so that trivially options C and B are incorrect.
For option $A)a{x^2} - 2hxy - b{y^2} = 0$after using the concept of $( - y){}^2 = - y \times ( - y) = {y^2}$so there is no possibility that square root will be negative and thus it is also incorrect.
If we directly substitute the value of $y = 0$then the answer is wrong because all other options are given as the image of the original value, suppose $y = 0$then we get \[a{x^2} + 2hxy + b{y^2} = 0 \Rightarrow a{x^2}\]is wrong (we only need to find the image not the values of the given equation)