A common tangent to the conics $x^2=6 y$ and $2 x^2-4 y^2=9$ is:
A common tangent to the conics $x^2=6 y$ and $2 x^2-4 y^2=9$ is:
$x-y=\frac{3}{2}$
$x+y=1$
$x+y=\frac{9}{2}$
$x-y=1$
Solution
$
\begin{aligned}
&x^2-6 y \\
&2 x^2-4 y^2=9
\end{aligned}
$
Consider the line,
$
x-y=\frac{3}{2}
$
On solving (i) and (iii), we get only
$
x=3, y=\frac{3}{2}
$
Hence $\left(3, \frac{3}{2}\right)$ is the point of contact of conic (i), and line (iii)
On solving (ii) and (iii), we get only $x=3$,
$
y=\frac{3}{2}
$
Hence $\left(3, \frac{3}{2}\right)$ is also the point of contact of conic (ii) and line (iii).
Hence line (iii) is the common tangent to both the given conics