Question

Line is $x+y=2$ is tangent to the curve $x^2=3-2y$ at the point then find point of the constant

Answer 1

Given: $x^2=3-2y.............\left(1\right)$

Tangent: $x+y=2............\left(2\right)$

Now, let the point of contact be $(x_1,y_2)$, so equation of tangent at $(x_1,y_2)$,

$=>x_{1}x=3-2(\frac{(y+y_1)}{2}$

$=>x_{1}x+y=3-y_1..............\left(3\right)$

Equation $\left(3\right)$ and $\left(2\right)$ represents the same So, comparing them we get,

$=>x_1=1,y_1=1$

So, point of contact is $(1,1)$.