Animated Solution for Mathematics - Conic Sections: The tangent to the parabola $y^2=4x$ at the point where it intersects the circle $x^2+y^2=5$ in the first quadrant, passes through the point :

We are given a parabola defined by the equation $y^2=4x$ and a circle defined by $x^2+y^2=5$. Our objective is to determine the equation of the tangent line at their point of intersection located in the first quadrant.

To find the intersection, we substitute the parabola's equation $y^2=4x$ into the circle's equation $x^2+y^2=5$. This yields the following quadratic equation:

Factoring the quadratic, we obtain:

This gives us two potential values for $x$: $x=-5$ and $x=1$. Since the parabola $y^2=4x$ requires $x\ge 0$ for real values of $y$, we discard $x=-5$ as an extraneous solution.

Setting $x=1$ in the parabola equation, we find $y^2=4$, which implies $y=\pm 2$. Because we are restricted to the first quadrant, we select the positive root, $y=2$. Thus, the point of intersection is $P(1,2)$.

The parabola is in the standard form $y^2=4ax$, where $a=1$. The equation of the tangent line to a parabola at a point $(x_1,y_1)$ is given by the formula:

Substituting our point $P(1,2)$ and the value $a=1$ into this formula, we get:

Dividing both sides by $2$, we arrive at the equation of the tangent line:

To verify the result, we test the point $(\frac{3}{4},\frac{7}{4})$ against our derived tangent line equation. Substituting $x=\frac{3}{4}$ into the equation $y=x+1$:

Since the calculated $y$ matches the given coordinate, the tangent line $y=x+1$ is confirmed as the correct solution.