Conics Half yearly Past Paper questions.... (1 Viewer)

kevda1st · Feb 21, 2009

These were in my schools half yearly extension 2 paper...there are no solutions....

1. X^2 +3Y^2 =1 (a) and 4X^2 +y^2 = 1 (b) are equations of 2 ellipses in the xy-plane.

I) show that the equation
X^2 +3Y^2 - 1 + K(4X^2 +y^2 - 1) =0 (*)
where k is a constant represents a curve passing trough the intersection of (a) and (b).

II) Find the values of k for which (*) repesents an ellipes

2. An ellise has the equation X^2/100 +Y^2/75 =1
III) find the equation of the circle that is tangential to the ellipse at P(5 ,7.5) and Q(5 , -7.5)

Trebla · Feb 21, 2009

kevda1st said:
These were in my schools half yearly extension 2 paper...there are no solutions....

1. X^2 +3Y^2 =1 (a) and 4X^2 +y^2 = 1 (b) are equations of 2 ellipses in the xy-plane.

I) show that the equation
X^2 +3Y^2 - 1 + K(4X^2 +y^2 - 1) =0 (*)
where k is a constant represents a curve passing trough the intersection of (a) and (b).

II) Find the values of k for which (*) repesents an ellipes

2. An ellise has the equation X^2/100 +Y^2/75 =1
III) find the equation of the circle that is tangential to the ellipse at P(5 ,7.5) and Q(5 , -7.5)

Q1
i)
$Conditions for point of intersection$\\x^2+3y^2=1 \Rightarrow x^2 = 1 - 3y^2\\4(1-3y^2)+y^2=1\Rightarrow y^2 = \dfrac{3}{11}\\\Rightarrow x^2 = \dfrac{2}{11}\\$Now consider $\\x^2+3y^2-1+k(4x^2+y^2-1)\\$Sub the values of $x^2$ and $y^2\\x^2+3y^2-1+k(4x^2+y^2-1) = \dfrac{2}{11}+\dfrac{9}{11}-1 + k(\dfrac{8}{11}+\dfrac{3}{11}-1)\\=0\\$Therefore point of intersection of both ellipses satisfies this curve$
ii)
$x^2+3y^2-1+k(4x^2+y^2-1)=0\\\Rightarrow (4k+1)x^2 + (k+3)y^2 = k + 1\\\Rightarrow \dfrac{4k+1}{k+1}x^2+\dfrac{k+3}{k+1}y^2 = 1\\$For ellipse we need $\dfrac{4k+1}{k+1} > 0$ and $\dfrac{k+3}{k+1} > 0 \\$ i.e. $ (4k+1)(k + 1) > 0 \Rightarrow k > -\dfrac{1}{4}$ or $k < -1\\$Also, $(k+3)(k+1) > 0 \Rightarrow k > -1$ or $k < -3\\$Satisfying intersection of all required conditions gives $k > -\dfrac{1}{4}$ and $k < -3$

kevda1st · Feb 21, 2009

for questions 2..what does tangential even mean??

cyl123 · Feb 21, 2009

tangential means the circle TOUCHES the ellipse at that point and share the same tangent

the easiest way i can think of to do the question is to find the normal of the ellipse at both points. Since the circle and ellipse share the same tangents, they share the same normal. knowing that normals to circle pass thru centre,, this can be used to find centre and then radius of circle

Aerath · Feb 21, 2009

I think it means that the circle just touches the ellipse, but doesn't cross it (ie, one point of intersection).

cyl123 · Feb 21, 2009

So the normal to ellipse (and circle) at (5,7.5) is 4x-2y=5
the normal at (5, -7.5) is 4x+2y=5
Since radius is perpendicular to tangent of circle, which is shared by ellipse, then the radius must be the normal to the ellipse, so where the 2 normals meet will be the centre.

Solving simulataneously gives x = 5/4, y=0

And radius is found using distance formula to be sqrt(70.3125)

So equation of circle is (x-5/4)^2+y^2 = 70.3125

Conics Half yearly Past Paper questions.... (1 Viewer)

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