Question

The equation of the common tangent to $x^2=6y$ and $2x^2-4y^2=9$ can be

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

Answer 1

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

Let $y=mx+c$ be the equation of the common tangent
Solving this with parabola $x^2=6y,$ we get
$\Rightarrow x^2=6(mx+c)\Rightarrow x^2-6mx-6c=0$
From condition of tangency, we get
$D=0\Rightarrow 36m^2+24c=0\Rightarrow c=-\frac{3}{2}{m}^2\dots \left(i\right)$
Equation tangent becomes
$y=mx-\frac{3}{2}{m}^2\Rightarrow 2y=2mx-3m^2$
This is also a tangent to $2x^2-4y^2=9$, so
$2x^2-(2mx-3m^2{)}^2=9\Rightarrow (2-4m^2)x^2+12m^{3}x-9m^4-9=0$
From condition of tangency, we get
$D=0\Rightarrow 144m^6-72(m^4+1)(2m^2-1)=0\Rightarrow 2m^6-(2m^6-m^4+2m^2-1)=0\Rightarrow m^4-2m^2+1=0\Rightarrow (m^2-1{)}^2=0\therefore m=\pm 1$

Hence, the required equation of tangents are
$x-y=\frac{3}{2}\text{and}x+y=-\frac{3}{2}$