Conic Sections: Tricky Questions Solved (3)

Question

Find the equation of the tangent to the parabola $y^2 = 8x$ at the point $(2, 4)$.

This looks like a one-step formula application, but JEE Advanced regularly trips students up by varying the parametric form. Let’s do it three ways and see which is fastest.

Solution — Step by Step

Why This Works

For any curve $f(x, y) = 0$, the tangent at $(x_1, y_1)$ uses the slope $dy/dx$ evaluated at that point. The slick “T = 0” form for conics — tangent equation $T = 0$ — comes from a clever substitution rule (e.g., $y^2 \to yy_1$, $x \to (x + x_1)/2$, etc.) that gives the tangent equation directly.

The substitution rule for $y^2 = 4ax$:

• $x \to (x + x_1)/2$
• $y^2 \to y y_1$
• $4ax \to 2a(x + x_1)$

So $y y_1 = 2a(x + x_1)$ becomes the tangent.

Alternative Method

Parametric form. For $y^2 = 4ax$, points on the parabola can be written as $(at^2, 2at)$. Tangent at parameter $t$: $y = x/t + at$ (or $ty = x + at^2$).

Here $a = 2$, $(at^2, 2at) = (2t^2, 4t) = (2, 4)$, so $t = 1$. Tangent: $y(1) = x + 2(1)^2 \implies y = x + 2$. ✓

Common Mistake

Students confuse the tangent formula for circles ($x x_1 + y y_1 = r^2$) with the parabola formula ($y y_1 = 2a(x + x_1)$). The replacement rule is conic-specific. If you don’t remember which is which, fall back to implicit differentiation — it always works.

Another trap: applying $y y_1 = 4a(x + x_1)$ instead of $y y_1 = 2a(x + x_1)$. The factor on $4a$ gets halved when you apply the "$x \to (x + x_1)/2$" rule. Slow down, write the substitution carefully, and you’ll get it right every time.