Find equation of tangent to parabola y²=8x at point (2,4)

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Question

Find the equation of the tangent to the parabola y2=8xy^2 = 8x at the point (2,4)(2, 4).

Solution — Step by Step

The standard form of a parabola is y2=4axy^2 = 4ax. Comparing with y2=8xy^2 = 8x:

4a=8    a=24a = 8 \implies a = 2

So this parabola has focus at (2,0)(2, 0) and directrix x=2x = -2.

We can verify that (2,4)(2, 4) lies on the parabola: 42=164^2 = 16 and 8×2=168 \times 2 = 16

For the parabola y2=4axy^2 = 4ax, the equation of the tangent at point (x1,y1)(x_1, y_1) is:

yy1=2a(x+x1)yy_1 = 2a(x + x_1)

Here a=2a = 2, x1=2x_1 = 2, y1=4y_1 = 4. Substituting:

y(4)=2(2)(x+2)y(4) = 2(2)(x + 2) 4y=4(x+2)4y = 4(x + 2) y=x+2y = x + 2

The equation of the tangent at (2,4)(2, 4) is:

y=x+2\boxed{y = x + 2}

or equivalently, xy+2=0x - y + 2 = 0.

Why This Works

The tangent formula yy1=2a(x+x1)yy_1 = 2a(x + x_1) is derived using implicit differentiation. Differentiating y2=4axy^2 = 4ax with respect to xx:

2ydydx=4a    dydx=2ay2y\frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}

At (2,4)(2, 4): slope =2×24=1= \frac{2 \times 2}{4} = 1.

Using point-slope form: y4=1(x2)    y=x+2y - 4 = 1(x - 2) \implies y = x + 2.

Both methods — the formula and implicit differentiation — give the same result. The formula is faster for board exams; the differentiation approach shows you understand the concept.

Alternative Method

Using implicit differentiation from scratch:

y2=8xy^2 = 8x

Differentiate both sides with respect to xx:

2ydydx=8    dydx=4y2y\frac{dy}{dx} = 8 \implies \frac{dy}{dx} = \frac{4}{y}

At point (2,4)(2, 4): slope m=4/4=1m = 4/4 = 1.

Tangent: y4=1(x2)    y=x+2y - 4 = 1(x - 2) \implies y = x + 2.

The tangent formula T=0T = 0 for a parabola is formed by replacing y2y^2 with yy1yy_1 and xx with x+x12\frac{x + x_1}{2} in the parabola equation. This is a systematic “T = 0” rule valid for all conics — learn it once and apply to circle, parabola, ellipse, hyperbola all in the same way.

Common Mistake

Students sometimes use the normal formula instead of the tangent formula, or forget the "x+x1x + x_1" form. The tangent at point (x1,y1)(x_1, y_1) on y2=4axy^2 = 4ax is yy1=2a(x+x1)yy_1 = 2a(x + x_1) — note it’s x+x1x + x_1 (not xx1x - x_1). This asymmetry catches many students off guard. Always derive it from the slope if you forget.

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