If the curves $2 x^2+3 y^2=6$ and $a x^2+4 y^2=4$ intersect orthogonally, then a =

2 1 3 none of these

2

We know that the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { and } \frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ intersect orthogonally iff $a^2-b^2=c^2-d^2$ Therefore, the curves $2 x^2+3 y^2=6$ and $a x^2+4 y^2=4$ will intersect orthogonally, if $3-2=\frac{4}{a}-1 \Rightarrow 2=\frac{4}{a} \Rightarrow a=2$