The equation of a tangent to the parabola, x² = 8y, which makes an angle θ with the positive directi

Q: Wait, why do we equate the derivative to $\tan \theta$?

Great question! The derivative $\frac{dy}{dx}$ gives the slope of the tangent line at any point on the curve. Geometrically, the slope of any line is also defined as the tangent of the angle it makes with the positive x-axis. Since both represent the slope of the same tangent line, they must be equal.

Q: Can we use the standard tangent formula $y = mx + c$ instead?

Absolutely! For a parabola $x^2 = 4ay$, the tangent with slope $m$ is $y = mx - am^2$. Here, $4a = 8$, so $a = 2$. Substituting $m = \tan \theta$ and $a = 2$, we get $y = x \tan \theta - 2 \tan^2 \theta$. Rearranging this will lead you to the same result. The method we used is just a more fundamental, first-principles approach.

Q: Why did we divide by $\tan \theta$ at the end?

We did that to match the format of the given options. The options are expressed in terms of $\cot \theta$. Since $\cot \theta = \frac{1}{\tan \theta}$, dividing the equation by $\tan \theta$ naturally transforms the terms into the required form.

Question

The equation of a tangent to the parabola, $x^2 = 8y$, which makes an angle $	heta$ with the positive direction of x-axis, is :

Options:

(A) $x = y\cot 	heta + 2	an 	heta$

(B) $x = y\cot 	heta - 2	an 	heta$

(C) $y = x	an 	heta - 2\cot 	heta$

(D) $y = x	an 	heta + 2\cot 	heta$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to explore the elegant intersection of geometry and calculus. We are tasked with finding the equation of a tangent to the parabola $x^2 = 8y$ that makes an angle $	heta$ with the positive $x$-axis.

Imagine this parabola as a smooth, symmetric bowl opening towards the sky. Our goal is to find the equation of a line that just grazes this bowl at a specific point, with its orientation perfectly defined by the angle $	heta$.

### The Bridge Between Calculus and Geometry

First, let us ground ourselves in the definitions. We know that the slope $m$ of any straight line is the tangent of the angle it makes with the positive $x$-axis. Thus, $m = 	an 	heta$. This is our geometric anchor.

Now, let us bring in the power of calculus. The slope of a tangent to a curve at any point is given by the derivative $\frac{dy}{dx}$. For our parabola $x^2 = 8y$, we differentiate both sides with respect to $x$:

$$
\frac{d}{dx}(x^2) = \frac{d}{dx}(8y)
$$

This gives us $2x = 8 \frac{dy}{dx}$. Solving for the slope, we find:

$$
\frac{dy}{dx} = \frac{x}{4}
$$

### The Synthesis of Ideas

This is where the magic happens. We have two expressions for the slope of the same tangent line: the geometric $m = 	an 	heta$ and the calculus-derived $\frac{dy}{dx} = \frac{x}{4}$.

By equating them, we find the $x$-coordinate of the point of contact:

$$
\frac{x}{4} = 	an 	heta \Rightarrow x = 4 	an 	heta
$$

Now that we have the horizontal position, we need the vertical one. Since the point of contact must lie on the parabola, we substitute $x = 4 	an 	heta$ into $x^2 = 8y$.

This yields $(4 	an 	heta)^2 = 8y$, or $16 	an^2 	heta = 8y$. Simplifying this, we get $y = 2 	an^2 	heta$. We have successfully pinpointed the exact location where our tangent touches the parabola: $(4 	an 	heta, 2 	an^2 	heta)$.

### Constructing the Final Equation

With the slope $m = 	an 	heta$ and the point of contact $(x_1, y_1) = (4 	an 	heta, 2 	an^2 	heta)$ in hand, we use the point-slope form of a line: $y - y_1 = m(x - x_1)$.

Substituting our values, we get:

$$
y - 2 	an^2 	heta = 	an 	heta (x - 4 	an 	heta)
$$

Expanding the right side, we have $y - 2 	an^2 	heta = x 	an 	heta - 4 	an^2 	heta$. Rearranging the terms, we arrive at:

$$
y = x 	an 	heta - 2 	an^2 	heta
$$

To align this with standard forms, we can rearrange the terms to isolate $x$. Finally, we obtain the **equation of the tangent**:

$$
x = y \cot 	heta + 2 	an 	heta
$$

You have just derived the equation of the tangent from first principles! Keep this logical flow in your toolkit—it will serve you well in many more complex problems.

Animated Solution for Mathematics - Conic Sections: The equation of a tangent to the parabola, $x^{2} = 8 y$ , which makes an angle $θ$ with the positive direction of x-axis, is :

Select Answer:

Visualized Solution

Visualizing the Parabola $x^{2} = 8 y$

The Tangent and its Angle $θ$

Defining the Slope $m$

Slope from Calculus $\frac{d y}{d x}$

Differentiating the Parabola

Isolating $\frac{d y}{d x}$

Equating the Two Slopes

Finding the $x$ -coordinate

Finding the $y$ -coordinate

Simplifying for $y$

Equation of the Tangent Line

Substituting into Point-Slope Form

Expanding the Equation

Rearranging the Terms

Matching with the Given Options

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Bridge Between Calculus and Geometry

The Synthesis of Ideas

Constructing the Final Equation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a tangent to the parabola, x2=8y, which makes an angle θ with the positive direction of x-axis, is :

Select Answer:

Visualized Solution

Visualizing the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2=8y

The Tangent and its Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Defining the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Slope from Calculus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​