# Conic Problems help me pleas...

• Jan 5th 2007, 09:44 PM
Conic Problems help me pleas...
Circles
Write the Eqn:
The circle is tangent to the line $\displaystyle 5x+12y = 26$ & the center is at the origin.
(I specifically dont knw where the circle is perpendicular to a part of this line)

Parabola
Write the Eqn:
Length of the latus rectum is 10 and the parabola opens upward.

Thanks in advance :)
• Jan 6th 2007, 12:13 AM
CaptainBlack
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Circles
Write the Eqn:
The circle is tangent to the line $\displaystyle 5x+12y = 26$ & the center is at the origin.
(I specifically dont knw where the circle is perpendicular to a part of this line)

The equation of the circle is:

$\displaystyle x^2+y^2=r^2$

rewrite the line as:

$\displaystyle y = (-5/12)x+13/6$

Now substiture this for $\displaystyle y$ in the equation of the circle:

$\displaystyle x^2+((-5/12)x+13/6)^2=r^2$

which simplifies down to:

$\displaystyle 169 x^2 - 260 x - 144 r^2 + 676=0$.

Now that the line is a tangent to the circle means that there is only one
point where they meet which corresponds to a double root of the quadratic.

The quadratic has a double root when the discriminant is zero, ie:

$\displaystyle 260^2-4\times 169 \times (-144 r^2+676)=0$

which is a quadratic in $\displaystyle r$. This has roots $\displaystyle r=-2, 2$,
and as $\displaystyle r$ must be positive the root we require is $\displaystyle r=2$.

So the equation of the circle is:

$\displaystyle x^2+y^2=4$

RonL
• Jan 6th 2007, 12:25 AM
CaptainBlack
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Circles
Parabola
Write the Eqn:
Length of the latus rectum is 10 and the parabola opens upward.

There is not quite enough information here, but if the vertex is at $\displaystyle (h,k)$ then the equation of the papabola is:

$\displaystyle 4p(y-k)=(x-h)^2$

and the latus rectum is $\displaystyle 4p$, so here we have:

$\displaystyle 10(y-k)=(x-h)^2$

For an explanation of this see here.

RonL
• Jan 6th 2007, 02:55 AM
ticbol
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Circles
Write the Eqn:
The circle is tangent to the line $\displaystyle 5x+12y = 26$ & the center is at the origin.
(I specifically dont knw where the circle is perpendicular to a part of this line)

Parabola
Write the Eqn:
Length of the latus rectum is 10 and the parabola opens upward.

Thanks in advance :)

Here is one way for the circle.

If the circle is tangent to the given line, then the radius of the circle touching the line at the point of tangency is perpendicular to the line. Hence this particular radius is the shortest line segment from the center of the circle to the given line. So this particular radius is the distance of the origin of the circle from the given line.

Given:
Center of circle = origin = (0,0)
Tangent line is 5x +12y = 26, or 5x +12y -26 = 0

Distance, d of (x1,y1) from line Ax +By +C = 0 is
d = | A*x1 +B*y1 +C | / sqrt(A^2 +B^2) ------------**

So,
r = |5*0 +12*0 -26| / sqrt(5^2 +12^2)
r = |-26| / sqrt(169)
r = 26 / 13
r = 2

The standard equation of a circle whose center is the origin is
x^2 +y^2 = r^2

Therefore, the circle is
x^2 +y^2 = 2^2
or, x^2 +y^2 = 4 ------------answer.

.................................................. ................

Question re parabola has not enough data to find the parabola.
• Jan 6th 2007, 08:36 AM
Soroban

Quote:

The circle is tangent to the line $\displaystyle 5x+12y = 26$ & the center is at the origin.
Write the equation of the circle.

The center is at the origin $\displaystyle (h,k) = (0,0)$
. . so the equation is: .$\displaystyle x^2 + y^2 \:=\:r^2$
Hence, we need only the radius $\displaystyle r.$

There is a formula for this situation.

The distance from a point $\displaystyle P(x_1,y_1)$ to the line $\displaystyle Ax + By + C \:=\:0$

. . is given by: .$\displaystyle d \;= \;\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+B^2}}$

So the distance from $\displaystyle C(0,0)$ to $\displaystyle 5x + 12y - 26 \:=\:0$ is:

. . $\displaystyle r \:=\:\frac{|5(0) + 12(0) - 26|}{\sqrt{5^2+12^2}} \;=\;\frac{26}{13} \:=\:2$

Therefore, the equation of the circle is: .$\displaystyle x^2 + y^2 \:=\:4$

• Jan 7th 2007, 05:51 AM
Hello
Thanks very much!
Sorry about the parabola..
ill add that the vertex was also at the origin
But kindly check my answer which is
$\displaystyle x^2=10y$
• Jan 7th 2007, 08:33 AM
CaptainBlack
Quote:

Originally Posted by ^_^Engineer_Adam^_^
Hello
Thanks very much!
Sorry about the parabola..
ill add that the vertex was also at the origin
But kindly check my answer which is
$\displaystyle x^2=10y$

Which is what you get if you put (0,0) in for the vertex in my post on this
problem.

RonL