Animated Solution for Mathematics - Conic Sections: Let $L$ be a tangent line to the parabola $y^2=4x-20$ at $(6,2)$. If $L$ is also a tangent to the ellipse $\frac{x^2}{2}+\frac{y^2}{b}=1$, then the value of $b$ is equal to :

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are witnessing a beautiful conversation between two fundamental conic sections: the parabola and the ellipse.

Imagine standing on a coordinate plane. You see a parabola, $y^2=4x-20$, curving gracefully to the right. You are handed a specific point on this curve, $P(6,2)$.

Our mission is to find a line $L$ that is tangent to this parabola at $P$ and then discover the hidden parameter $b$ that allows this same line to perfectly graze an ellipse, $\frac{x^2}{2}+\frac{y^2}{b}=1$.

When we need the tangent to a conic at a specific point $(x_1,y_1)$, we don't need to resort to calculus or complex derivatives. We have a secret weapon: the $T=0$ transformation. This method is a cornerstone of coordinate geometry.

For any conic, we replace $y^2$ with $y{y}_1$ and $x$ with $\frac{x+x_1}{2}$. Applying this to our parabola $y^2=4x-20$ at the point $(6,2)$, we substitute $y_1=2$ and $x_1=6$:

Look at how the equation simplifies. The $4$ divided by $2$ becomes $2$, and distributing that $2$ gives us $2x+12$. Subtracting $20$ leaves us with $2y=2x-8$.

Dividing by $2$, we arrive at the elegant equation of our line $L$:

This is the line that bridges our two worlds. Its slope $m$ is $1$, and its y-intercept $c$ is $-4$.

Now, we introduce the ellipse: $\frac{x^2}{2}+\frac{y^2}{b}=1$. The problem tells us that our line $L$ is also tangent to this ellipse.

For a line $y=mx+c$ to be tangent to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, it must satisfy the condition:

This condition is the mathematical manifestation of the line having exactly one point of contact with the ellipse. If the discriminant of the resulting quadratic equation were anything other than zero, the line would either miss the ellipse or cut through it at two points. We want that perfect, single-point kiss.

We have everything we need. We know $m=1$, $c=-4$, and from our ellipse equation, $a^2=2$. The term $b^2$ in our formula is simply $b$ in our given equation.

Let's plug these into our condition:

Calculating the squares, we get $16=2+b$. The path to the solution is now clear. Subtracting $2$ from $16$, we find:

And there it is! The value of $b$ that allows the line $L$ to be tangent to the ellipse is $14$. It is a moment of pure mathematical harmony—the line $L$, born from the parabola, finds its perfect place against the ellipse. Keep practicing this visualization; it is the key to mastering coordinate geometry!