Question

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

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

Answer 1

Answer: (A)

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

Answer 2

The correct answer is (A) : $x-y=\frac{3}{2}$
Let $y=(mx+c)$ is tangent to $x^2=6y$
Now , $x^2=6(mx+c)$
So, $x^2-6mx-6c=0$
Put $D=b^2-4ac=0$
$⇒ c=\frac{-3}{2}m^2$
$\therefore$ we get $y=mx-\frac{3}{2}m^2$ $.....(1)$
and given hyperbola equation is $2x^2-4y^2=9$
$⇒\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{9}{4}}=1$ $....(2)$
Since, equation (1) is a tangent of equation (2) then $c^2=a^2m^2-b^2$
$⇒\frac{9}{4}m^4=9m^2-\frac{9}{4}$
$⇒m^4=2m^2-1$
$⇒m^4-2m^2+1=0$
$⇒(m^2-1)^2=0$
⇒$$m=±1
Therefore , for m=1 , equation of tangent is $x-y=\frac{3}{2}$