Animated Solution for Mathematics - Conic Sections: If the tangent to the parabola $y^2=x$ at a point $(\alpha ,\beta ),(\beta >0)$ is also a tangent to the ellipse, $x^2+2y^2=1$, then $\alpha$ is equal to :

Imagine you are standing on a coordinate plane, looking at two elegant curves: a parabola $y^2=x$ and an ellipse $x^2+2y^2=1$. They seem separate, yet they are connected by a single, shared line—a common tangent.

This is the heart of our problem. We are not just solving for $\alpha$; we are uncovering the hidden geometric relationship between these two shapes.

Let us start with the parabola $y^2=x$. We are given a point $(\alpha ,\beta )$ on this curve. Since the point lies on the parabola, it must satisfy the equation, giving us the crucial constraint:

We are told $\beta >0$, which places our point in the first quadrant. Now, we need the equation of the tangent line at this point.

Using the standard formula for a tangent to $y^2=4ax$ at $(x_1,y_1)$, which is $y{y}_1=2a(x+x_1)$, we identify $4a=1$, so $a=1/4$. The tangent equation becomes:

To connect this tangent to the ellipse, we need it in the slope-intercept form $y=mx+c$. Substituting $\alpha ={\beta }^2$ into our tangent equation, we get $y\beta =\frac{1}{2}(x+{\beta }^2)$.

Dividing by $\beta$, we find:

Here, the slope $m$ is $\frac{1}{2\beta }$ and the intercept $c$ is $\frac{\beta }{2}$. This line is the key to everything.

Now, we turn our attention to the ellipse $x^2+2y^2=1$. In standard form, this is $\frac{x^2}{1}+\frac{y^2}{1/2}=1$, where $a^2=1$ and $b^2=1/2$.

A line $y=mx+c$ is tangent to this ellipse if and only if $c^2=a^2{m}^2+b^2$. This is the powerful condition we have been waiting for. Substituting our values, we get:

Now, let us simplify this. We have:

To clear the denominators, we multiply the entire equation by $4{\beta }^2$, resulting in ${\beta }^4=1+2{\beta }^2$, or:

This is a quadratic in disguise! Let $t={\beta }^2$. Then $t^2-2t-1=0$.

Using the quadratic formula:

Since ${\beta }^2$ must be positive, we reject $1-\sqrt{2}$ and accept ${\beta }^2=1+\sqrt{2}$. Finally, since $\alpha ={\beta }^2$, we have:

The beauty of this problem lies in how the two curves, through the shared tangent, dictate the exact position of the point $(\alpha ,\beta )$. Keep practicing this, and you will see the elegance in every coordinate geometry problem you face!