Answer
$-1$
The equations $2x^2 +3y^2=5$ and $y^2=x^3$ are perpendicular to each other at the point $(1,1)$.
Work Step by Step
We have $2x^2 +3y^2=5 ...(a)\\ y^2=x^3 ....(b)$
So equation (a) becomes: $2x^2 +3(x^3) =5$ and $y^2=x^3=(1)^3 =1$
The slope of the equation is: $2x^2 +3y^2=5$.
$4x+6y \dfrac{dy}{dt}=0 $
or, $\dfrac{dy}{dt} =\dfrac{-2x}{3y}$
Now, $m_1 (1,1)=\dfrac{dy}{dt}(1,1) =\dfrac{-2(1)}{3(1)}=\dfrac{-2}{3}$
and $m_2 (1,1)= \dfrac{3x^2}{2y}(1,1) =\dfrac{3}{2}$
$m_1 \times m_2= \dfrac{-2}{3} \times \dfrac{3}{2} =-1$
So, the equations $2x^2 +3y^2=5$ and $y^2=x^3$ are perpendicular to each other at the point $(1,1)$.
