Show that the equation of the tangent to the parabola $y^2=4ax$ at the point (p,q) is $qy=2a(x+p)$

Question

Question: Show that the equation of the tangent to the parabola $y^2=4ax$ at the point (p,q) is $qy=2a(x+p)$

These are my two approaches:

First approach:

If we have $(p,q)$ as $(x_1,y_1)$

$$y^2=4ax$$

$$ 2y \left( \frac{dy}{dx}\right) = 4a$$

$$ \frac{dy}{dx} = \frac{2a}{y} $$

At $y=q$

$$ \frac{dy}{dx} = \frac{2a}{q} $$

$$ m = \frac{2a}{q} $$

As equation of tangent is

$$ (y-y_1) = m(x-x_1) $$

Plugging in values

$$ (y-q) = \frac{2a}{q}(x-p) $$

$$ y-q = \frac{2ax}{q} - \frac{2ap}{q} $$

$$ y-q = \frac{2ax-2ap}{q} $$

$$ qy - q^2 = 2ax-2ap $$

$$ qy -q^2 = 2a(x-p) $$

But this does not satisfy the proof

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My second approach:

If $(p,q)$ are points on $y^2=4ax$ then

plugging into $$ y^2 = 4ax $$

at $x=p$

$$ y^2 = 4ap $$

$$ y = \sqrt{4ap} $$

at $y=q$

$$ q^2 = 4ax $$

$$ x = \frac{q^2}{4a} $$

Finding gradient:

$$y^2=4ax$$

$$ 2y \left( \frac{dy}{dx}\right) = 4a$$

$$ \frac{dy}{dx} = \frac{2a}{y} $$

At $y = \sqrt{4ap} $

$$ \frac{dy}{dx} = \frac{a}{\sqrt{ap}} $$

$$ (y-y_1) = m(x-x_1) $$

$$ y-\sqrt{4ap}=\frac{a}{\sqrt{ap}}(x-\frac{q^2}{4a})$$

But I don't think this gives me the proof as well...

Answer 1

In the first approach, $$qy-q^2=2a(x-p)$$ But $(p,q)$ is a point on the parabola $y^2=4ax$.

Thus, $$\tag1q^2=4ap$$ Substituting, $$qy-4ap=2ax-2ap$$ $$qy=2a(x+p)$$

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In the second approach, $$y-\sqrt{4ap}=\frac{a}{\sqrt{ap}}(x-\frac{q^2}{4a})$$ From $(1)$, $$\tag2\frac{q^2}{4a}=p$$ $$\tag3\sqrt{ap}=\frac{q}{2}$$ $$\tag4\sqrt{4ap}=q$$ Using $(2),(3)$ and $(4)$, $$y-q=\frac{2a}{q}(x-p)$$ $$qy-q^2=2a(x-p)$$ which is the same equation as in the first case and reduces to $$qy=2a(x+p)$$

Answer 2

With your first approach is easier, I think. You got $\;y'=\frac{2a}y\;$ , so at point $\;(p,q)\;$ the slope is $\;\frac{2a}q\;$ , and thus the line with this slope and through this point is

$$y-q=\frac{2a}q(x-p)\iff qy-q^2=2a(x-p)$$

but $\;q^2=4ap\;$ since $\;(p,q)\;$ is on the parabola, and then:

$$qy-4ap=2ax-2ap\implies qy=2ax+2ap=2a(x+p)$$

Answer 3

My approach:

We have the equation of parabola : $y^2 = 4ax$

By taking the derivative of the above equation to get its gradient $m$, we have

$2y \left( \frac{dy}{dx}\right) = 4a$

Rearranging the above equation,

$\left(\frac{dy}{dx}\right)$ = $\frac{4a}{2y}$ = $m$

The equation of tangent is : $(y-y_1) = m(x-x_1)$

Putting in value of $m$ here, we have

$(y-y_1) = \frac{4a}{2y} (x-x_1)$

Also we know that $(y_1,x_1)=(p,q)$. So substituting the values we get

$(y-p) = \frac{4a}{2y} (x-q)$

Now by multiplying $2y$ on both sides we get

$2y^2 - 2qy = 4ax - 4ap$

$2y^2 - 4ax + 4ap = 2qy$

Now here we can substitute the value of $y^2$ with $4ax$ (From the equation of the parabola)

By doing so we have,

$8ax - 4ax + 4ap = 2qy$

$4ax + 4ap = 2qy$

$4a(x + p) = 2qy$

$\color{green}{qy=2a(x+p)}$