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

The Correct Option is A

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 \(x-y=\frac{3}{2}\)