Question

Given below are Matching type questions, with two columns(each having some items) each.Each item of 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 items 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 $△$, where $△=abc+2fgh-a{f}^2-b{g}^2-c{h}^2$ or $\left(\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right)$
Now, match the entries from the following two columns

Answer 1

(A)$\because \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$
On squaring both sides, we get
$\frac{x}{a}+\frac{y}{b}+2\sqrt{\frac{xy}{ab}}=1$
or $\frac{x}{a}+\frac{y}{b}-1=2\sqrt{\frac{xy}{ab}}$
Again squaring both sides, we get
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+1+\frac{2xy}{ab}-\frac{2y}{b}-\frac{2x}{a}=4\left(\frac{xy}{ab}\right)$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}-2\left(\frac{xy}{ab}\right)+1-\frac{2y}{b}-\frac{2x}{a}=0$ .......$\left(1\right)$
Now, comparing $\left(1\right)$ with $A{x}^2+2Hxy+B{y}^2+2Gx+2Fy+C=0$ then
$A=\frac{1}{a^2},H=-\frac{1}{ab},B=\frac{1}{b^2},G=-\frac{1}{a},F=-\frac{1}{b},C=1$
On substituting the above values in the below condition,
$△=ABC+2FGH-A{F}^2-B{B}^2-C{H}^2$
$=\frac{1}{a^2{b}^2}-\frac{2}{a^2{b}^2}-\frac{1}{a^2{b}^2}-\frac{1}{a^2{b}^2}-\frac{1}{a^2{b}^2}$
$=\frac{-4}{a^2{b}^2}\ne 0$
and $H^2=AB$
$\because △\ne 0,H^2=AB$
$\therefore$ conic is non-degenerate and a parabola.
(B)Now comparing $3x^2+10xy+3y^2-15x-21y+18=0$
with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ we get
$a=3,h=5,b=3,g=\frac{-15}{2},f=\frac{-21}{2},c=18$ ,
On substituting the above values in below condition, we get
$=△=abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$=3\times 3\times 18+2\times \frac{-21}{2}\times \frac{-15}{2}\times 5-3\times {\left(\frac{-21}{2}\right)}^2-3\times {\left(\frac{-15}{2}\right)}^2-18\times {\left(5\right)}^2$
$=162+288-450=450-450=0$
and $h^2\ne ab$
$\because △=0,h^2\ne ab$ the conic is degenerate and a pair of intersecting straight lines.
(C)Now comparing $8x^2-4xy+5y^2-16x-14y+17=0$ with
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ we get
$a=8,h=-2,b=5,g=-8,f=-7,c=17$
On substituting the above values in below condition, we get
$△=abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$=8\times 5\times 17+2\times -7\times -8\times -2-8\times {(-7)}^2-5\times {(-8)}^2-17\times {(-2)}^2$
$=255-224-147-320-68=-504\ne 0$
and $h^2<ab$
$\therefore △\ne 0$ and $h^2<ab$ then conic is non-degenerate and an ellipse.