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A common tangent to the conics $x^2=6 y$ and $2 x^2-4 y^2=9$ is:
Options:
Solution:
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Verified Answer
The correct answer is:
$x-y=\frac{3}{2}$
$x-y=\frac{3}{2}$
$$
\begin{aligned}
&x^2-6 y \\
&2 x^2-4 y^2=9
\end{aligned}
$$
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.
\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.
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