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Chapter 14: Problem 314

Find the equation of the line that is tangent to the parabola \(y^{2}=2 p x\) at the point \(\left(x_{1}, y_{1}\right)\)

Short Answer

Expert verified
The equation of the tangent line to the parabola \(y^2 = 2px\) at the point \((x_1, y_1)\) is given by: \[y_1y - y_1^2 = p(x - x_1)\]

Step by step solution

01

Find the derivative of the function

To find the slope of the tangent line at a given point, we need the first derivative of the function. The original equation of the parabola is given by \(y^2 = 2px\). Differentiating both sides of the equation with respect to x, we get: \[\frac{d(y^2)}{dx} = \frac{d(2px)}{dx}\] Applying the chain rule on the left side, we have: \[2y\frac{dy}{dx} = 2p\] Now, solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{p}{y}\]
02

Find the slope at the given point

Now, we will find the slope of the tangent line at the given point \((x_1, y_1)\). To do this, plug in the given coordinates into the derivative we found in the last step: \[\left. \frac{dy}{dx} \right|_{(x_1, y_1)} = \frac{p}{y_1}\]
03

Use point-slope form to find the equation of the tangent line

To find the equation of the tangent line, we will use the point-slope form of the equation of a line, which is: \[y - y_1 = m(x - x_1)\] Here, \(m\) is the slope of the tangent line at the given point, which we found in step 2. Plugging in the values, we get: \[y - y_1 = \frac{p}{y_1}(x - x_1)\]
04

Simplify the equation of the tangent line

Finally, we will simplify the equation we found in step 3. To do this, we need to eliminate the fraction. Multiply both sides by \(y_1\), we get: \[y_1(y - y_1) = p(x - x_1)\] Expanding the left side, we get: \[y_1y - y_1^2 = p(x - x_1)\] So, the equation of the tangent line to the parabola \(y^2 = 2px\) at the point \((x_1, y_1)\) is: \[y_1y - y_1^2 = p(x - x_1)\]

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Most popular questions from this chapter

Find the points of contact of the horizontal and vertical tangents to \(\mathrm{x}=2-3 \sin \theta, \mathrm{y}=3+2 \cos \theta\)

Find the equations of the tangent and the normal to the curve: \(\mathrm{y}=\mathrm{x}^{2}+4 \mathrm{x}+2\), at a point where the tangent is also perpendicular to the line \(2 \mathrm{x}-4 \mathrm{y}+5=0\)

Find the points where the curve described by the following parametric equations has a zero slope: \(x=3 t /\left(1+t^{3}\right), y=3 t^{2} /\left(1+t^{3}\right)\)

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Find the equation of the curve passing through \((1,2)\) which has the property that its slope at any point is equal to twice the abscissa of the point.
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