Question

Find the length of latus rectum of the parabola $\left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) = {\left( {bx + ay - ab} \right)^2}$
A) $\dfrac{{ab}}{{\sqrt {{a^2} + {b^2}} }}$
B) $\dfrac{{2ab}}{{\sqrt {{a^2} + {b^2}} }}$
C) $a + b$
D) $a + b + \dfrac{{ab}}{{\sqrt {ab} }}$

Answer 1

We already know that the latus rectum is perpendicular to the axis line that passes through the focus of a parabola. We'll use the standard parabola equation in this case. The parabola's vertex, focus, and directrix are all included in the standard equation. We can simplify the equation and clearly state the length of the latus rectum by arranging the data in standard form.

Complete step-by-step solution:
Given equation,
$\left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) = {\left( {bx + ay - ab} \right)^2}$
The standard form of the parabola can be written as
${(y - k)^2} = \pm 4p(x - h)$
Where $(y - k)$ gives is the distance from the axis, $4p$ gives the latus rectum and $\left( {x - h} \right)$ gives the distance from tangent at vertex.
So, we will arrange the given equation into the standard form by opening the brackets of LHS first as
$\Rightarrow \left( {{a^2}{x^2} + {b^2}{x^2} + {a^2}{y^2} + {b^2}{y^2}} \right) = {\left( {bx + ay - ab} \right)^2}$
Now opening the brackets of RHS as well, we get
$\Rightarrow \left( {{a^2}{x^2} + {b^2}{x^2} + {a^2}{y^2} + {b^2}{y^2}} \right) = \left( {{b^2}{x^2} + {a^2}{y^2} + {a^2}{b^2} + 2abxy - 2{a^2}by - 2a{b^2}x} \right)$
In the given equation ${b^2}{x^2}$ and ${a^2}{y^2}$ get canceled because they are on both sides. Therefore, we get
$\Rightarrow \left( {{a^2}{x^2} + {b^2}{y^2}} \right) = \left( {{a^2}{b^2} + 2abxy - 2{a^2}by - 2a{b^2}x} \right)$
Now taking $- 2abxy$ on the LHS, we get
$\Rightarrow \left( {{a^2}{x^2} + {b^2}{y^2} - 2abxy} \right) = \left( {{a^2}{b^2} - 2{a^2}by - 2a{b^2}x} \right)$
We can further write this equation in the form of squares on LHS as
$\Rightarrow {\left( {ax - by} \right)^2} = \left( {{a^2}{b^2} - 2{a^2}by - 2a{b^2}x} \right)$ $\left[ {\because {x^2} + {y^2} - 2xy = {{\left( {x - y} \right)}^2}} \right]$
Now taking ab common from the RHS, we get
$\Rightarrow {\left( {ax - by} \right)^2} = ab\left( {ab - 2ay - 2bx} \right)$
By multiplying and dividing $({a^2} + {b^2})$ on the LHS, we get
$\Rightarrow ({a^2} + {b^2}){\left( {\dfrac{{ax - by}}{{\sqrt {{a^2} + {b^2}} }}} \right)^2} = ab\left( {ab - 2ay - 2bx} \right)$
And multiplying and dividing $2\sqrt {{a^2} + {b^2}}$ on the RHS, we get
$\Rightarrow ({a^2} + {b^2}){\left( {\dfrac{{ax - by}}{{\sqrt {{a^2} + {b^2}} }}} \right)^2} = ab \times 2\sqrt {{a^2} + {b^2}} \left( {\dfrac{{ab - 2ay - 2bx}}{{\sqrt {{{(2a)}^2} + {{(2b)}^2}} }}} \right)$
By taking $({a^2} + {b^2})$ from LHS to RHS, we get
$\Rightarrow {\left( {\dfrac{{ax - by}}{{\sqrt {{a^2} + {b^2}} }}} \right)^2} = ab \times \dfrac{{2\sqrt {{a^2} + {b^2}} }}{{({a^2} + {b^2})}}\left( {\dfrac{{ab - 2ay - 2bx}}{{\sqrt {{{(2a)}^2} + {{(2b)}^2}} }}} \right)$
$\Rightarrow {\left( {\dfrac{{ax - by}}{{\sqrt {{a^2} + {b^2}} }}} \right)^2} = \dfrac{{2ab}}{{\sqrt {{a^2} + {b^2}} }}\left( {\dfrac{{ab - 2ay - 2bx}}{{\sqrt {{{(2a)}^2} + {{(2b)}^2}} }}} \right)$
Taking negative sign common from the RHS, we get
$\Rightarrow {\left( {\dfrac{{ax - by}}{{\sqrt {{a^2} + {b^2}} }}} \right)^2} = - \dfrac{{2ab}}{{\sqrt {{a^2} + {b^2}} }}\left( {\dfrac{{2ay + 2bx - ab}}{{\sqrt {{{(2a)}^2} + {{(2b)}^2}} }}} \right)$
Now by comparing the above equation with the standard equation
Latus Rectum $= \dfrac{{2ab}}{{\sqrt {{a^2} + {b^2}} }}$
Hence, the correct option is B.

Note: Here we will consider $\left( {\dfrac{{ax - by}}{{\sqrt {{a^2} + {b^2}} }}} \right)$ as the distance from the axis and $\left( {\dfrac{{2ay + 2bx - ab}}{{\sqrt {{{(2a)}^2} + {{(2b)}^2}} }}} \right)$ as the distance from tangent at vertex. Always remember that the axis and the tangent at vertex are perpendicular lines to get a better understanding. Here we have to always simplify the given equation in standard form of parabola.