Finding the locus of Q

**free_to_fly** · December 19th 2007, 02:24 AM

Given that S (square root 3,0) is a focus of the ellipse with equation 3x^2+4y^2=36. The origin is O and P is any point on the ellipse. A line is drawn from O perpendicular to the tangent to the ellipse at P and this line meets the line SP, produced if necessary, in the point Q. Show that the locus of Q is a circle.

Can someone show me the diagram for this (I think my diagram's wrong) and give me a hint as to where to start? Then hopefully I should be able to get the answer. Thanks.

**earboth** · December 19th 2007, 05:37 AM

Originally Posted by free_to_fly

Given that S (square root 3,0) is a focus of the ellipse with equation 3x^2+4y^2=36. The origin is O and P is any point on the ellipse. A line is drawn from O perpendicular to the tangent to the ellipse at P and this line meets the line SP, produced if necessary, in the point Q. Show that the locus of Q is a circle.

Can someone show me the diagram for this (I think my diagram's wrong) and give me a hint as to where to start? Then hopefully I should be able to get the answer. Thanks.

Hi,

here is the diagram. All those black dots belong to the locus. You easily can see that actually the result should be a circle.

The following is a suggestion how to get the equation of the locus:

1. derivate the equation of the ellipse implicitely
2. choose a point P on the ellipse and calculate the slope of the tangent
3. calculate the perpendicular direction to this slope
4. calculate the equation of a line through the origin with the perpendicular slope
5. calculate the equation of a line through the focus S and the point P
6. calculate the coordinates of the intersection point
7. eliminate the parameter (probably the x-coordinate of point P)

Only for confirmation: The equation of this circle is: $(x-\sqrt{3})^2+y^2=12$ . That means the focus S is the center of the circle and the radius has the length $\sqrt{12}$

**free_to_fly** · December 19th 2007, 10:05 AM

When you say: "choose a point P on the ellipse", do you mean use the general point P being (acost, bsint)? I used that and got really complicated expressions for x and y at the point of intersection, so I must have done it wrong.

**earboth** · December 19th 2007, 09:01 PM

Originally Posted by free_to_fly

When you say: "choose a point P on the ellipse", do you mean use the general point P being (acost, bsint)? I used that and got really complicated expressions for x and y at the point of intersection, so I must have done it wrong.

Hello,

probably I've done your problem in a very complicated way but it works:

$3x^2+4y^2=36~\implies~y=\pm \frac12 \cdot \sqrt{36-3x^2}$

The arbitrary point P which is placed on the ellipse becomes $P\left(p, \frac12 \cdot \sqrt{36-3p^2}\right)$

The slope of the tangent in P is calculated by:

Implicite derivation of the ellipse: $6x+8y \cdot y'=0~\implies~ y'=-\frac{3x}{4y}$ Therefore the slope of the tangent is $m_t=-\frac{3p}{2 \cdot \sqrt{36-3p^2}}$

The perpendicular direction is $m=\frac{2 \cdot \sqrt{36-3p^2}}{3p}$ and the equation of the line through the origin and perpendicular to the tangent becomes:

$\boxed{y = \frac{2 \cdot \sqrt{36-3p^2}}{3p} \cdot x}$

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

The line connecting S and P can be calculated by the 2-point-formula:

$\frac{y-0}{x-\sqrt{3}} =\frac{\frac12 \cdot \sqrt{36-3p^2} - 0}{p-\sqrt{3}}$ . Solve for y to get the equation of this line:

$\boxed{y=\frac{(\sqrt{36-3p^2}) \cdot (x-\sqrt{3})}{2 \cdot (p-\sqrt{3})}}$

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

If there exists en intersection point the RHSs of the 2 equation must be equal:

$\frac{2 \cdot \sqrt{36-3p^2}}{3p} \cdot x = \frac{(\sqrt{36-3p^2}) \cdot (x-\sqrt{3})}{2 \cdot (p-\sqrt{3})}$ . Solve for x. After a lot of work and at least three very silly mistakes you'll get:

$x = \frac{3 \cdot \sqrt{3} \cdot p}{4 \cdot \sqrt{3} - p}$ [1]

Plug in this value into the equation of the first line to calculate the y-coordinate of the intersection point:

$y = \frac{6 \cdot \sqrt{12-p^2} \cdot p}{4 \cdot \sqrt{3} - p}$ [2]

Thus the intersection point is $C \left(\frac{3 \cdot \sqrt{3} \cdot p}{4 \cdot \sqrt{3} - p}~,~\frac{6 \cdot \sqrt{12-p^2} \cdot p}{4 \cdot \sqrt{3} - p} \right)$

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

To calculate the locus I've solved the equation [1] for p and plugged this term into the equation [2] :

$p = \frac{4 \cdot \sqrt{3} \cdot x}{x+3 \cdot \sqrt{3}}$ and therefore:

$y = \frac{6 \cdot \sqrt{12-\left(\frac{4 \cdot \sqrt{3} \cdot x}{x+3 \cdot \sqrt{3}} \right)^2} \cdot \left(\frac{4 \cdot \sqrt{3} \cdot x}{x+3 \cdot \sqrt{3}} \right)}{4 \cdot \sqrt{3} - \left(\frac{4 \cdot \sqrt{3} \cdot x}{x+3 \cdot \sqrt{3}} \right)}$ By first sight you won't believe it but you can simplify this monster equation to:

$y = \pm \sqrt{-x^2 + 2 \cdot \sqrt{3} \cdot x +9}~\implies~y^2 = -x^2 + 2 \cdot \sqrt{3} \cdot x +9$ This equation can be transformed to the result I gave you in my previous post.

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