Question

If graph of \[xy = 1\] is reflected in \[y = 2x\] to give the graph\[12{x^2} + rxy + s{y^2} + t = 0\] then
A.\[r = 7\]
B.\[s = - 12\]
C.\[t = 25\]
D.\[r + s = 19\]

Answer 1

Hint: Here we need to find the values of the unknowns. We will first assume a point on the curve and then we will find its image point about the line. We will use the fact that the midpoint of a point and its image point about the given line will lie on the same line. Further, we will find the midpoint and we will put the value of abscissa and ordinate of midpoint in the equation of the line. We will also use these two points to find the slope. After solving these two equations, we will get the required equation.

Complete step-by-step answer:
Let the point on the graph be \[A\left( {\alpha ,\beta } \right)\] and let the image of the point A about the line\[y = 2x\] be \[B\left( {a,b} \right)\].
Midpoint of AB \[\left( {\dfrac{{a + \alpha }}{2},\dfrac{{b + \beta }}{2}} \right)\]
The midpoint of point A and B will lie on the line \[y = 2x\].
We will put the value of abscissa and ordinate of that midpoint in the equation of the line.
2.\[\dfrac{{a + \alpha }}{2} = \dfrac{{b + \beta }}{2}\]
After simplification, we get
$\Rightarrow$ \[\beta + b = 2a + 2\alpha \]……………….\[\left( 1 \right)\]
The slope of line joining the points A and B is equal to the slope of the line which is perpendicular to the \[y = 2x\]
Therefore, slope of line AB \[ = - \dfrac{1}{2}\]
We can also write the slope of line AB as \[\dfrac{{\beta - b}}{{\alpha - a}}\]
Therefore,
$\Rightarrow$ \[\dfrac{{\beta - b}}{{\alpha - a}} = - \dfrac{1}{2}\]
Cross multiplying the fractions, we get
$\Rightarrow$ \[\beta - b = \dfrac{1}{2}a - \dfrac{1}{2}\alpha \]………………\[\left( 2 \right)\]
Subtracting equation 2 from equation 1, we get
$\Rightarrow$ \[2b = \dfrac{3}{2}a + \dfrac{5}{2}\alpha \]
We will find the value of \[\alpha \] in terms of a and b.
Therefore,
$\Rightarrow$ \[\alpha = \dfrac{{4b - 3a}}{5}\]
We will put the value of \[\alpha \] in equation 1.
$\Rightarrow$ \[\beta + b = 2a + 2\left( {\dfrac{{4b - 3a}}{5}} \right)\]
Multiplying the terms, we get
$\Rightarrow$ \[\beta + b = 2a + \dfrac{{8b - 6a}}{5}\]
Subtracting and adding like terms, we get
$\Rightarrow$ \[\beta = \dfrac{{3b + 4a}}{5}\]
Since point\[\left( {\alpha ,\beta } \right)\] lies on the curve \[xy = 1\]
Therefore,
$\Rightarrow$ \[\left( {\dfrac{{4b - 3a}}{5}} \right)\left( {\dfrac{{3b + 4a}}{5}} \right) = 1\]
Simplifying the equation, we get
$\Rightarrow$ \[12{a^2} - 7ab - 12{b^2} + 25 = 0\]
We will replace \[a\] and \[b\] with \[x\] and \[y\] to express the equation in general form.
$\Rightarrow$ \[12{x^2} - 7xy - 12{y^2} + 25 = 0\]
Comparing the given equation \[12{x^2} + rxy + s{y^2} + t = 0\] with the obtained equation $\Rightarrow$ \[12{x^2} - 7xy - 12{y^2} + 25 = 0\], we get
\[ \Rightarrow r = - 7\\
\Rightarrow s = - 12\\
\Rightarrow t = 25\\
\Rightarrow s + r = - 19
\]
Therefore, the correct options are B and C.

Note: Here we have used an image of a point about the given line. An image of a point about the given line is at the same distance from the line and the midpoint of a point and its image point lies on the same line.
We need to remember that the point that lies on the line or any curve will satisfy the equation of the given line and curve.