Question

The equation $3x^2+4y^2-18x+16y+43=k$ represents

• A. an empty set, if $k<0$
• B. an ellipse, if $k>0$
• C. a point, if $k=0$
• D. cannot represent a real pair of straight lines for any value of $k$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A an ellipse, if $k>0$
B an empty set, if $k<0$
C cannot represent a real pair of straight lines for any value of $k$
D a point, if $k=0$

Given, an equation as $3x^2+4y^2-18x+16y+43=k$

$\Rightarrow 3(x^2-6x)+4(y^2+4y)=k-43$

$\Rightarrow 3(x^2-6x+9)+4(y^2+4y+4)=(k-43)+27+16$

$\Rightarrow 3(x-3{)}^2+4(y+2{)}^2=k$

$\Rightarrow \frac{(x-3{)}^2}{4}+\frac{(y+2{)}^2}{3}=\frac{k}{12}$

For $k<0$, RHS will be negative which is impossible.

$\therefore$ For $k<0$, it represents empty set

Consider,

$\frac{(x-3{)}^2}{4}+\frac{(y+2{)}^2}{3}=\frac{k}{12}$

$\Rightarrow \frac{(x-3{)}^2}{\frac{k}{3}}+\frac{(y+2{)}^2}{\frac{k}{4}}=1$

For equation to be ellipse, $\frac{k}{3}>0,\frac{k}{4}>0\Rightarrow k>0$

If $k=0$,

$\Rightarrow \frac{(x-3{)}^2}{4}+\frac{(y+2{)}^2}{3}=\frac{0}{12}$

$\Rightarrow \frac{(x-3{)}^2}{4}+\frac{(y+2{)}^2}{3}=0$

Sum of the squares $=0$, only if individual terms are zero.

$\Rightarrow (x-3)=0,(y+2)=0\Rightarrow (x,y)=(3,-2)$ which represents a point.

Consider the given equation, $3x^2+4y^2-18x+16y+43=k$ and

compare with the general equation of pair of straight lines as $A{x}^2+B{y}^2+2Hxy+2Gx+2Fy+C=0$

represents only if $H^2=AB$

$\Rightarrow A=3,B=4,H=0$

But, $H^2=0,AB=3\times 4=12$

Since $H^2\ne AB$, for any value of $k$ the given equation does not represent a pair of straight lines.

$\therefore$ All options are true.