Question

Number of intersecting points of the conics $4x^2+9y^2=1$ and $4x^2+y^2=4$ is

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

Answer: (D)

$0$

For curve I:
$4x^2+9y^2=1\Rightarrow \frac{x^2}{{\left(\frac{1}{2}\right)}^2}+\frac{y^2}{{\left(\frac{1}{3}\right)}^2}=1$
Which is an equation of ellipse with $a=\frac{1}{2}$ and $b=\frac{1}{3}$

For curve II:

$4x^2+y^2=4$
$\Rightarrow \frac{x^2}{1}+\frac{y^2}{4}=1$
$\Rightarrow \frac{x^2}{(1{)}^2}+\frac{y^2}{(2{)}^2}=1$
Which is also an equation of ellipse with $a=1$ and $b=2$.
Hence, there is no intersecting point.