Animated Solution for Mathematics - Conic Sections: If the line $2x+\sqrt{6}y=2$ touches the hyperbola $x^2-2y^2=4$, then the point of contact is

We are tasked with finding the point of contact where the line $2x+\sqrt{6}y=2$ is tangent to the hyperbola $x^2-2y^2=4$. In coordinate geometry, this point of contact represents the unique intersection where the line and the curve share a single point.

Before proceeding, we must express the hyperbola in its standard form, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Starting with the given equation $x^2-2y^2=4$, we divide the entire equation by $4$:

This simplifies to the standard form:

From this, we identify our parameters as $a^2=4$ and $b^2=2$.

To find the point of contact $(x_1,y_1)$, we utilize the powerful $T=0$ method. For a hyperbola, the equation of the tangent at a point $(x_1,y_1)$ is obtained by replacing $x^2$ with $x{x}_1$ and $y^2$ with $y{y}_1$.

The general formula is:

Substituting our known values for $a^2$ and $b^2$, the theoretical equation of the tangent becomes:

We now compare our theoretical equation with the given line $2x+\sqrt{6}y=2$. To align the constants, we divide the given line by $2$:

Since both equations represent the same line, their coefficients must be proportional. Comparing the coefficients of $x$:

Comparing the coefficients of $y$:

Solving for $y_1$, we find:

By transforming the problem into a comparison of coefficients, we have bypassed complex algebraic substitution. The point of contact is: