Find the equation of the tangent to the parabola y^2 = 5x which is parallel to the line y = 4x+1 . Also find the point of contact?

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Steve M Share
Dec 23, 2017

The tangent to the Parabola that is parallel to $y = 4 x + 1$ is:

$y = 4 x + \frac{5}{16}$

Which meets the Parabola at the coordinate:

$\left(\frac{5}{64} , \frac{5}{8}\right)$

Explanation:

We have a parabola given by:

${y}^{2} = 5 x$

graph{y^2=5x [-5, 5, -5, 5]}

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So if we differentiate the parabola equation (implicitly) we have:

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 5 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{2 y}$

Comparing the given line equation $y = 4 x + 1$ with the standard line equation $y = m x + x$ we see that its gradient is given by:

$m = 4$

So we seek a tangent equation for the parabola with the same slope, thus we require:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \implies \frac{5}{2 y} = 4 \implies y = \frac{5}{8}$

When $y = \frac{5}{8}$ the corresponding $x$-coordinate is given by:

${y}^{2} = 5 x \implies 5 x = \frac{25}{64} \implies x = \frac{5}{64}$

So the point of contact is $\left(\frac{5}{64} , \frac{5}{8}\right)$. So, using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the tangent equations is;

$y - \frac{5}{8} = 4 \left(x - \frac{5}{64}\right)$
$\therefore y - \frac{5}{8} = 4 x - \frac{5}{16}$
$\therefore y = 4 x + \frac{5}{16}$

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