Find the area bounded by function $y^2=16-2x$, the tangent to the curve at the point $(6,2)$ and the $y$-axis

First we find a tangent line of function $y^2=16-2x$ at $T(6,2)$:

$y_t-y_0=f'(x_0)(x-x_0)$ where $x_0=6,y_0=2$

$y^2=16-2x\Rightarrow y=\sqrt{16-2x}$ or $y=-\sqrt{16-2x}$

Derivative of $y=\sqrt{16-2x}$ at $x_0=6$ is $f'(x_0)=\frac{-1}{\sqrt{16-2x_0}}=-\frac{1}{2}\Rightarrow$ tangent of $f(x)$ at $T(6,2)$ is $y_t=-\frac{1}{2}x+5$.

The total area bounded by $f(x),T$ and $y$ axis is the difference between the are below tangent line $y_t$ and below $f(x)$.

$A=A_t-A_f$

$$A_t=\int\limits_0^6 y_t\mathrm dx=\int\limits_0^6 \left(-\frac{1}{2}x+5\right)\mathrm dx=\int\limits_0^6 -\frac{1}{2}x\mathrm dx+\int\limits_0^6 5\mathrm dx=21$$

How to find are below $y^2=16-2x$?

• I don't understand the boundary. What does $T(6,2)$ mean? Because including just the point $(6,2)$ has no affect on the shape of the boundary. – Elliot G Feb 9 '16 at 9:12
• @Elliot G $T(6,2)$ is the point in the first quadrant where tangent line of $y=\sqrt{16-2x}$ is. If you graph the function $y^2=16-2x$ and its tangent at $T(6,2)$, the bounded area of $y^2=16-2x,T,y$ axis has three points: $T(6,2),A(0,4),B(0,5)$. – user300048 Feb 9 '16 at 9:18

Hint:

Since you have the upper ($y>0$) half parabola you can integrate the function $$y=\sqrt{16-2x}$$ so the area is: $$A_f=\int_0^6\sqrt{16-2x}dx$$ that, with the substitution $16-2x=t \rightarrow -2dx=dt$ becomes:

$$A_f=\int_6^4\sqrt{t}dx$$

can you do from this?

You can find the area under the curve: Pink area $A_1 =\int\limits_0^6 \sqrt{16-2x} \mathrm {dx}$. You can do this using integration by substitution.

You already have the area of the trapezium, so just need to subtract $A_1$ from $21$ to find the required area.

Or you can imagine the curve turned the other way and find the area between the curve and the $y$-axis.

Pink area $A_2=\int\limits_2^4 x \mathrm {dy}= \int\limits_2^4 \frac {16-y^2}2 \mathrm {dy}$. This is an easier integral.

To find the orange area required you will need to subtract $A_2$ from the area of the triangle.