Conic Sections: Speed-Solving Techniques (8)

Question

Find the equation of the tangent to the parabola $y^2 = 8x$ at the point $(2, 4)$. Also, find the equation of the normal at the same point. JEE Main 2023 standard.

Solution — Step by Step

Why This Works

The “T = 0” form (chord of contact) is one of the fastest techniques in conic sections. For any conic $S(x, y) = 0$, the tangent at point $(x_1, y_1)$ is obtained by replacing $x^2 \to xx_1$, $y^2 \to yy_1$, $x \to (x+x_1)/2$, $y \to (y+y_1)/2$.

For the parabola $y^2 = 4ax$: $y^2$ becomes $yy_1$, $x$ becomes $(x+x_1)/2$, giving $yy_1 = 4a \cdot (x+x_1)/2 = 2a(x+x_1)$. That’s where the formula comes from.

Alternative Method

Differentiate implicitly: $2y\dfrac{dy}{dx} = 8$, so $\dfrac{dy}{dx} = 4/y = 4/4 = 1$ at $(2, 4)$. Then the tangent is $y - 4 = 1 \cdot (x - 2)$, i.e., $y = x + 2$. Same result.

The differentiation method works for any curve; the T=0 trick is faster for standard conics.