Question

Given below are matching type questions, with two columns (each having some items) each. Each item of the column$I$ has to be matched with the items of column$II$, by enclosing the correct match.
Note: An item of column$I$ can be matched with more than one item of column$II$. All the items of column$II$ have to be matched.
The equation of conics represented by the general equation of second degree $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ and the discriminant of above equation is represented by $\Delta$, where $\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}$ or $\left| \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right|$.
Now match the entries from the following two columns

(A). $\text{ }The\text{ }conic\text{ }is\text{ }represented\text{ }by\text{ }the\text{ }equation\text{ }\sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=1\text{ } is \left( a\ne 0,b\ne 0 \right)$	(i)$\text{Degenerate}$
(B). $\text{ The conic is represented by the equation } 3{{x}^{2}}+10xy+3{{y}^{2}}-15x-21y+18=0$	(ii)$\text{Non-Degenerate}$
(C).$\text{The conic is representedby the equation } 8{{x}^{2}}-4xy+5{{y}^{2}}-16x-14y+17=0$	(iii)$\text{ A parabola}$

Answer 1

We start solving the problems by converting the given conics into the standard form as given in the form. We then find the values of a, b, c, h, f, g, and find the discriminant of the given conic using the formula given in the problem to check the degeneracy of the conic. We then compare the values of ${{h}^{2}}$ and $ab$ to know what is the conic that the given equation is representing which ultimately gives us the required answer.

Complete step-by-step solution
According to the problem, we need to match the two columns given in the table.
Let us first check for (A), we need to check the properties of the conic whose equation is $\sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=1$, $\left( a\ne 0,b\ne 0 \right)$.
Let us first conic in terms of the standard form $A{{x}^{2}}+2Hxy+B{{y}^{2}}+2Gx+2Fy+C=0$.
Let us square the given equation of conic on both sides.
$\Rightarrow {{\left( \sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}} \right)}^{2}}={{1}^{2}}$.
$\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}+2\sqrt{\dfrac{xy}{ab}}=1$.
$\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}-1=-2\sqrt{\dfrac{xy}{ab}}$.
Let us square this equation on both sides.
$\Rightarrow {{\left( \dfrac{x}{a}+\dfrac{y}{b}-1 \right)}^{2}}={{\left( -2\sqrt{\dfrac{xy}{ab}} \right)}^{2}}$.
$\Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}+1-2\dfrac{x}{a}-2\dfrac{y}{b}+2\dfrac{xy}{ab}=4\dfrac{xy}{ab}$.
$\Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}-2\dfrac{xy}{ab}+\dfrac{{{y}^{2}}}{{{b}^{2}}}-2\dfrac{x}{a}-2\dfrac{y}{b}+1=0$.
Comparing with standard equation of conic $A{{x}^{2}}+2Hxy+B{{y}^{2}}+2Gx+2Fy+C=0$ we get $A=\dfrac{1}{{{a}^{2}}}$, $H=\dfrac{-1}{ab}$, $B=\dfrac{1}{{{b}^{2}}}$, $G=\dfrac{-1}{a}$, $F=\dfrac{-1}{b}$, $C=1$.
Now let us find the discriminant of this conic.
$\Rightarrow \Delta =ABC+2FGH-A{{F}^{2}}-B{{G}^{2}}-C{{H}^{2}}$.
$\Rightarrow \Delta =\left( \left( \dfrac{1}{{{a}^{2}}} \right)\times \left( \dfrac{1}{{{b}^{2}}} \right)\times \left( 1 \right) \right)+\left( 2\times \left( \dfrac{-1}{b} \right)\times \left( \dfrac{-1}{a} \right)\times \left( \dfrac{-1}{ab} \right) \right)-\left( \left( \dfrac{1}{{{a}^{2}}} \right)\times {{\left( \dfrac{-1}{b} \right)}^{2}} \right)-\left( \left( \dfrac{1}{{{b}^{2}}} \right)\times {{\left( \dfrac{-1}{a} \right)}^{2}} \right)-\left( 1\times {{\left( \dfrac{-1}{ab} \right)}^{2}} \right)$.
$\Rightarrow \Delta =\dfrac{1}{{{a}^{2}}{{b}^{2}}}-\dfrac{2}{{{a}^{2}}{{b}^{2}}}-\dfrac{1}{{{a}^{2}}{{b}^{2}}}-\dfrac{1}{{{a}^{2}}{{b}^{2}}}-\dfrac{1}{{{a}^{2}}{{b}^{2}}}$.
$\Rightarrow \Delta =-\dfrac{4}{{{a}^{2}}{{b}^{2}}}\ne 0$.
We know that if $\Delta \ne 0$, then the given conic is non-degenerate.
Let us consider ${{H}^{2}}$.
$\Rightarrow {{H}^{2}}={{\left( \dfrac{-1}{ab} \right)}^{2}}$.
$\Rightarrow {{H}^{2}}=\dfrac{1}{{{a}^{2}}{{b}^{2}}}$.
$\Rightarrow {{H}^{2}}=\left( \dfrac{1}{{{a}^{2}}} \right)\times \left( \dfrac{1}{{{b}^{2}}} \right)$.
$\Rightarrow {{H}^{2}}=AB$.
We know that if $\Delta \ne 0$ and ${{H}^{2}}=AB$ for a conic, then the given conic is a parabola.
∴ (A) matches with $(ii)$ and $(iii)$.
Let us first check for (B), we need to check the properties of the conic whose equation is $3{{x}^{2}}+10xy+3{{y}^{2}}-15x-21y+18=0$.
Comparing with standard equation of conic $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ we get $a=3$, $h=5$, $b=3$, $g=\dfrac{-15}{2}$, $f=\dfrac{-21}{2}$, $c=18$.
Now let us find the discriminant of this conic.
$\Rightarrow \Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}$.
$\Rightarrow \Delta =\left( 3\times 3\times 18 \right)+\left( 2\times \left( \dfrac{-21}{2} \right)\times \left( \dfrac{-15}{2} \right)\times 5 \right)-\left( 3\times {{\left( \dfrac{-21}{2} \right)}^{2}} \right)-\left( 3\times {{\left( \dfrac{-15}{2} \right)}^{2}} \right)-\left( 18\times {{5}^{2}} \right)$.
$\Rightarrow \Delta =162+\dfrac{1575}{2}-\dfrac{1323}{4}-\dfrac{675}{4}-450$.
$\Rightarrow \Delta =\dfrac{3150-1323-675}{4}-288$.
$\Rightarrow \Delta =\dfrac{1152}{4}-288$.
$\Rightarrow \Delta =288-288$.
$\Rightarrow \Delta =0$.
We know that if $\Delta =0$, then the given conic is degenerate.
Let us consider ${{h}^{2}}$.
$\Rightarrow {{h}^{2}}={{5}^{2}}$.
$\Rightarrow {{h}^{2}}=25$.
Let us consider $ab$.
$\Rightarrow ab=3\times 3$.
$\Rightarrow ab=9$.
$\Rightarrow {{h}^{2}}\ne ab$.
We know that if $\Delta =0$ and ${{h}^{2}}\ne ab$ for a conic, then the given conic is a pair of straight lines.
∴ (B) matches with $(i)$.
Let us first check for (C), we need to check the properties of the conic whose equation is $8{{x}^{2}}-4xy+5{{y}^{2}}-16x-14y+17=0$.
Comparing with standard equation of conic $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ we get $a=8$, $h=-2$, $b=5$, $g=-8$, $f=-7$, $c=17$.
Now let us find the discriminant of this conic.
$\Rightarrow \Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}$.
$\Rightarrow \Delta =\left( 8\times 5\times 17 \right)+\left( 2\times \left( -7 \right)\times \left( -8 \right)\times \left( -2 \right) \right)-\left( 8\times {{\left( -7 \right)}^{2}} \right)-\left( 5\times {{\left( -8 \right)}^{2}} \right)-\left( 17\times {{\left( -2 \right)}^{2}} \right)$.
$\Rightarrow \Delta =680-224-392-320-68$.
$\Rightarrow \Delta =-324\ne 0$.
We know that if $\Delta \ne 0$, then the given conic is non-degenerate.
Let us consider ${{h}^{2}}$.
$\Rightarrow {{h}^{2}}={{\left( -2 \right)}^{2}}$.
$\Rightarrow {{h}^{2}}=4$.
Let us consider $ab$.
$\Rightarrow ab=8\times 5$.
$\Rightarrow ab=40$.
$\Rightarrow {{h}^{2}} < ab$.
We know that if $\Delta \ne 0$ and ${{h}^{2}} < ab$ for a conic, then the given conic is an ellipse.
Option (C) matches with $(ii)$.

Note: We can also find the discriminant using the given determinant which will be easier to calculate if there are no variables present in the equation of the conic. We should know that degenerate conics can be written as a product of two linear equations whereas the non-degenerate conics cannot be written as that product. Similarly, we can also expect problems to find the value of the unknowns present in the given equation of the conic using the discriminant and other conditions.