# find the equation of the lines.

• Jul 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)

thank you
• Aug 1st 2012, 10:03 AM
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)

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.
• Aug 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
• Sep 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?
• Sep 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)

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.
• Sep 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