For which value of $p$ do the curves $2x^{2} + 3y^{2} =1$ and $px^{2} + 6y^{2} =1$ intersect orthogonally?
(Options:) $1/3$, $4$, $3$, $2/3$.
I have calculated $\text{grad } f$ for these two curves
\begin{align}
\text{grad } f_1 &= (4x, 6y)\\
\text{grad } f_2 &= (2px,12y) .
\end{align}
How do I proceed next? Thanks.
Best Answer
Hints:
$$\begin{align}&\nabla f=(4x\,,\,6y)\\{}\\&\nabla g=(2px\,,\,12y)\end{align}$$
and now the curves are orthogonal at their intersection point $\;(a,b)\;$ (assuming they're differentiable there and etc.) iff
$$\nabla f(a,b)\cdot\nabla g(a,b)=0$$
Well, do the math now with the dot product after you find the curves' intersection point $\;(a,b)\;$...or otherwise.