# Find eq. to ellipse

• Nov 3rd 2006, 11:17 AM
viet
Find eq. to ellipse
Find the equation of the line that is tangent to the ellipse
b^2x^2 + a^2y^2 = a^2b^2 in the first quadrant and forms with the coodiante axes the triangle with smallest possible area (a and b are positive constants)

im not sure how to do this problem, the teacher explained it but it was very confusing.
• Nov 3rd 2006, 12:52 PM
earboth
Quote:

Originally Posted by viet
Find the equation of the line that is tangent to the ellipse
b^2x^2 + a^2y^2 = a^2b^2 in the first quadrant and forms with the coodiante axes the triangle with smallest possible area (a and b are positive constants)

im not sure how to do this problem, the teacher explained it but it was very confusing.

Hello, viet,

have a look here: http://www.mathhelpforum.com/math-he...html#post26090

EB
• Nov 3rd 2006, 01:05 PM
Soroban
Hello, viet!

Quote:

Find the equation of the tangent to the ellipse [1] $b^2x^2 + a^2y^2 \:= \:a^2b^2$ in quadrant 1
which forms with the coodiante axes the triangle with smallest possible area.
( $a$ and $b$ are positive constants.)

Here is a little-known (but very convenient) formula . . .

The equation of the tangent to the ellipse: . $\frac{x^2}{a^2} + \frac{y^2}{b^2}\;=\;1$ at the point $(h,k)$
. . is given by: . $\frac{hx}{a^2} +\frac{ ky}{b^2}\:=\:1$

Then the intercepts of this tangent are: . $\left(\frac{a^2}{h},\,0\right)$ and $\left(0,\,\frac{b^2}{k}\right)$

The area of the triangle is: . $A \:=\:\frac{1}{2}\left(\frac{a^2}{h}\right)\left(\f rac{b^2}{k}\right) \;= \;\frac{a^2b^2}{2hk}$

For convenience, let $x = h,\:y = k.$ . Then we have: . $A \;= \;\frac{a^2b^2}{2}x^{-1}y^{-1}$

Differentiate: . $A' \;= \;\frac{a^2b^2}{2}\bigg[\left(x^{-1}\right)\left(\text{-}y^{-2}\right)y' + \left(\text{-}x^{-2}\right)\left(y^{-1}\right)\bigg]$

Equate to zero: . $-\frac{y'}{xy^2} - \frac{1}{x^2y} \;= \;0$

Multiply by $-x^2y^2:\;\;xy' + y \;=\;0\quad\Rightarrow\quad y \:=\:-xy'$ [2]

. . The ellipse is: .[1] $b^2x^2 + a^2y^2\:=\:a^2b^2$
. . Differentiate implicitily: . $2b^2x + 2a^2yy' \:=\:0\quad\Rightarrow\quad y' \:=\:-\frac{b^2x}{a^2y}$ [3]

Substitute [3] into [2]: . $y \:=\:-x\left(-\frac{b^2x}{a^2y}\right) \quad\Rightarrow\quad a^2y^2\:=\:b^2x^2$ [4]

Substitute [4] into [1]: . $b^2x^2 + b^2x^2\:=\:a^2b^2\quad\Rightarrow\quad 2x^2 - a^2\quad\Rightarrow\quad x = \frac{a}{\sqrt{2}}$

Substitute into [1] and get: . $y\,=\,\frac{b}{\sqrt{2}}$

Hence, the intercepts of the tangent are:
. . $\left(\frac{a^2}{\frac{a}{\sqrt{2}}},\:0\right) = (\sqrt{2}a,\,0)$ and $\left(0,\:\frac{b^2}{\frac{b}{\sqrt{2}}}\right) = (0,\,\sqrt{2}b)$

And I'll let you write the equation of that tangent line . . .

• Nov 3rd 2006, 08:53 PM
viet
i got it, thanks for your help