Question

A common tangent to the conics $x^2=6y$ and $2x^2-4y^2=9$ is :

• A. $x+y=\frac{9}{2}$
• B. $x+y=1$
• C. $x-y=\frac{3}{2}$
• D. $x-y=1$

Answer 1

The correct option is C $x-y=\frac{3}{2}$

$x^2-6y=4ay$

$4a=6$

$a=\frac{3}{2}$

Tangent is $x=\frac{y}{t}+at$

$tx=y+\frac{3}{2}{t}^2$

For $2x^2-4y^2=9$

$\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{9}{4}}=1$

$y=mx+\sqrt{(a^2{m}^2-b^2)}$

$y=mx+\sqrt{\frac{9m^2}{2}-\frac{9}{4}}$

Both the equations represent the same line

$m^4-2m^2+1=0$

$m^2=1$

$y=tx-\frac{3}{2}{t}^2$

$t=1$

$y=x-\frac{3}{2}$

$x-y=\frac{3}{2}$