> For the complete documentation index, see [llms.txt](https://yingshaoxo.gitbook.io/university-notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://yingshaoxo.gitbook.io/university-notes/high-level-math/gaokao/tangent-line.md).

# Tangent Line

**Tangent**: In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point.

![](/files/-Lf5honMwSGj0agEhKrb)

### key concept 1

The equation for the slope of the tangent line to $$f(x)$$ is $$f^\prime(x)$$.

$$f^\prime(x)$$ is the derivative of $$f(x)$$.

You can always get the slope of the tangent line by using $$f^\prime(x)$$.

> `f(x)` is nothing but a function or formula.

### key concept 2

And a line equation could be represent as:

* the `slope-intercept formula`: $$y = mx + b$$ (Where $$m$$ is the slope of the line, and $$b$$ is the y-intercept)

> The equation of any straight line, called a `linear equation`, can be written as: `y = mx + b`, where `m` is the `slope of the line` and `b` is the `y-intercept`.

* the `point-slope formula`: $$y - y\_1 = m(x - x\_1)$$ (This formula uses a point on the line, denoted by $$(x\_1, y\_1)$$, and the slope of the line, denoted by $$m$$, to calculate the slope-intercept formula for the line)

> `Point-slope` is the general form `y-y₁=m(x-x₁)` for linear equations. It emphasizes the `slope of the line` and `a point on the line`.

## Question and Answers

### `Area of a triangle` is $$18$$. The triangle is enclosed by a `Cartesian coordinate system` and `a tangent`. The equation of the tangent to the curve $$y = x^{-\frac{1}{2}}$$ is at the point $$(a, a^{-\frac{1}{2}})$$. What's the value of $$a$$?

> The answer is `64`

#### we should try to understand the question sentence by sentence first.

**1.Area of a triangle is :**

![](/files/-Lf5honYZyQRM-ZNbnfc)

**2. The triangle is enclosed by a Cartesian coordinate system and a tangent.**

* `tangent` is a line
* the `triangle` was created by the `Cartesian coordinate system` and `the tangent line`

**3. The equation of the tangent to the curve  is at the point .**

The slope of the tangent line is:

$$
\begin{align\*}
\text{Let's say } f(x) &= x^{-\frac{1}{2}}
\ \\
slope = f^\prime(x) &= -\frac{1}{2}x^{-\frac{1}{2}-1}
\\
&= -\frac{1}{2}x^{-\frac{3}{2}}
\ \\
slope = f^\prime(a) &= -\frac{1}{2}a^{-\frac{3}{2}}
\end{align\*}
$$

Now we already have a point and a slope, then we can generate `the equation of the line`:

$$
\begin{align\*}
y - y\_1 &= m(x - x\_1)
\ \\
y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}}
)(x - a)
\end{align\*}
$$

If we could get the x, y in the following picture, we can find an equation of `the area of the triangle`:

![](/files/-Lf5honcOdKqr-octWPO)

$$
\begin{align\*}
y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a)
\ \\
\text{when x = 0: } &
\\
y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(0 - a)
\\
y - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(- a)
\\
y - a^{-\frac{1}{2}} &= \frac{1}{2}a^{-\frac{3}{2} + 1}
\\
y - a^{-\frac{1}{2}} &= \frac{1}{2}a^{-\frac{1}{2}}
\\
y &= \frac{1}{2}a^{-\frac{1}{2}} + a^{-\frac{1}{2}}
\\
y &= \frac{3}{2}a^{-\frac{1}{2}}

\ \\
\text{when y = 0: } &
\\
0 - a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a)
\\

* a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}})(x - a)
  \\
* a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) - (-\frac{1}{2}a^{-\frac{3}{2}} \cdot a)
  \\
* a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{3}{2}} \cdot a)
  \\
* a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{3}{2} + 1})
  \\
* a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x) + (\frac{1}{2}a^{-\frac{1}{2}})
  \\
* a^{-\frac{1}{2}} - (\frac{1}{2}a^{-\frac{1}{2}}) &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x)
  \\
  (-1 - \frac{1}{2})a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x)
  \\
  (- \frac{3}{2})a^{-\frac{1}{2}} &= (-\frac{1}{2}a^{-\frac{3}{2}} \cdot x)
  \\
* \frac{3}{2}a^{-\frac{1}{2}} &= -\frac{1}{2}a^{-\frac{3}{2}} \cdot x
  \\
  \frac{- \frac{3}{2}a^{-\frac{1}{2}}}{-\frac{1}{2}a^{-\frac{3}{2}}} &= x
  \\
  (-\frac{3}{2} \cdot -\frac{2}{1})\frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x
  \\
  (-\frac{3}{2} \cdot -2)\frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x
  \\
  3 \cdot \frac{a^{-\frac{1}{2}}}{a^{-\frac{3}{2}}} &= x
  \\
  3 \cdot a^{(-\frac{1}{2}) - (-\frac{3}{2})} &= x
  \\
  3 \cdot a^{(-\frac{1}{2}) + (\frac{3}{2})} &= x
  \\
  3 \cdot a^1 &= x
  \\
  3 \cdot a &= x
  \\
  3a &= x

\ \\
\text{area of triangle} &= \frac{1}{2} \cdot height \cdot width
\\
18 &= \frac{1}{2} \cdot y \cdot x
\\
18 &= \frac{1}{2} \cdot \frac{3}{2}a^{-\frac{1}{2}} \cdot 3a
\\
18 &= (\frac{1}{2} \cdot \frac{3}{2} \cdot 3) \cdot a^{-\frac{1}{2}} \cdot a
\\
18 &= (\frac{9}{4}) \cdot a^{-\frac{1}{2}} \cdot a
\\
18 &= (\frac{9}{4}) \cdot a^{-\frac{1}{2} + 1}
\\
18 &= (\frac{9}{4}) \cdot a^{\frac{1}{2}}
\\
18 \cdot \frac{4}{9} &= a^{\frac{1}{2}}
\\
8 &= a^{\frac{1}{2}}
\\
8 &= \sqrt{a}
\\
8^2 &= (\sqrt{a})^2
\\
64 &= a
\\
a &= 64
\\
\end{align\*}
$$

So, finally, we have got the value of a, which is `64`.


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter, and the optional `goal` query parameter:

```
GET https://yingshaoxo.gitbook.io/university-notes/high-level-math/gaokao/tangent-line.md?ask=<question>&goal=<endgoal>
```

`ask` is the immediate question: it should be specific, self-contained, and written in natural language.
`goal` is optional and describes the broader end goal you are ultimately trying to accomplish on behalf of the user. GitBook uses it to tailor the answer towards what is most useful for that goal.

The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.