find the equation of the lines.

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July 30th 2012, 11:26 PM
rcs

find the equation of the lines.

there are two lines through the given point, which are tangent to the given curve. Find the equation each of these lines.

4x^2- 5xy + 2y^2 + 3x - 2y = 0 , P ( 2,3)

need assistance on this please.

thank you
August 1st 2012, 10:03 AM
earboth

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Re: find the equation of the lines.

Quote:

Originally Posted by rcs

there are two lines through the given point, which are tangent to the given curve. Find the equation each of these lines.

4x^2- 5xy + 2y^2 + 3x - 2y = 0 , P ( 2,3)

need assistance on this please.

thank you

1. All lines passing through P have the equation: $y = m(x-2)+3$

2. Replace the term y in your equation of the ellipse by m(x-2)+3. You'll get a quadratic in x:

$x^2(2m^2 - 5m + 4) - x(8m^2 - 20·m + 12) + 4(2m^2 - 5·m + 3) = 0$

Solve for x.

3. Usually a straight line intercepts an ellipse in 2 or in 1 point or it is a passante that means there are no common points. The case that there exists only one point of interception occurs if the straight line is a tangent to the ellipse. This will happen if the discriminant equals zero:

$\sqrt{- 2·m^2 + 5·m - 3 - 2·m^2 + 5·m - 3)} = 0$

Solve for m. You should come out with $m = 1~\vee~m=\frac32$

4. Plug in these values into the equation of the line.
August 2nd 2012, 01:53 AM
rcs

Re: find the equation of the lines.

thank you for enlightening my brain... millions of thanks sir.
God Bless
September 3rd 2012, 07:04 PM
rcs

Re: find the equation of the lines.

Quote:

Originally Posted by earboth

1. All lines passing through P have the equation: $y = m(x-2)+3$

2. Replace the term y in your equation of the ellipse by m(x-2)+3. You'll get a quadratic in x:

$x^2(2m^2 - 5m + 4) - x(8m^2 - 20·m + 12) + 4(2m^2 - 5·m + 3) = 0$

Solve for x.

3. Usually a straight line intercepts an ellipse in 2 or in 1 point or it is a passante that means there are no common points. The case that there exists only one point of interception occurs if the straight line is a tangent to the ellipse. This will happen if the discriminant equals zero:

$\sqrt{- 2·m^2 + 5·m - 3 - 2·m^2 + 5·m - 3)} = 0$

Solve for m. You should come out with $m = 1~\vee~m=\frac32$

4. Plug in these values into the equation of the line.

sir i just dont get when you said to use the y as m(x-2)+3 to ellipse equation?... but what is the ellipse formula sir?
September 3rd 2012, 10:34 PM
earboth

Re: find the equation of the lines.

Quote:

Originally Posted by rcs

there are two lines through the given point, which are tangent to the given curve. Find the equation each of these lines.

4x^2- 5xy + 2y^2 + 3x - 2y = 0 , P ( 2,3)

need assistance on this please.

thank you

Quote:

Originally Posted by rcs

sir i just dont get when you said to use the y as m(x-2)+3 to ellipse equation?... but what is the ellipse formula sir?

The marked equation describes an ellipse, not the ellipse.

From the attachment of my first post you can see that you have a "tilted" ellipse, that means the axes are not parallel to the coordinate axes, but the curve is still an ellipse.
September 4th 2012, 04:16 AM
rcs

Re: find the equation of the lines.

Quote:

Originally Posted by earboth

The marked equation describes an ellipse, not the ellipse.

From the attachment of my first post you can see that you have a "tilted" ellipse, that means the axes are not parallel to the coordinate axes, but the curve is still an ellipse.

i got it sir... i solved it manually and i was able to get that equation equal to 0. something that stuck me how you got square root (-2m^2 +5m -3 -2m^2 +5m - 3) = 0... is that the discriminant sir? which being plugged in by the values of the equation above?

thanks