<p>this one is from barron's:
if the graph of 3x^2 + 4y^2 -6x + 8y - 5 + 0 and ( x - 2 )^2 = 4( y + 2 ) are drawn on the same co -ordinate system at how many points do they intersect?
(a)0
(b)1
(c)2
(d)3
(e)4
how do i do this sum with only a scientific calci..
my tutor showed me the method of plotting the two equations and then see where the lines intersect but it takes too longg...
givin it this october... aint there any faster way to go about such sums?</p>
<p>It helps a lot to have a graphing calc. Otherwise, the only way would be to solve it algebraically..</p>
<p>You may not need to solve it exactly, but it will still take some time without a graphing calculator (as gxing indicated).</p>
<p>You can rewrite 3x^2 + 4y^2 -6x + 8y - 5 = 0
as 3(x^2 -2x +1) -3 + 4(y^2 + 2y + 1) -4 -5 = 0
or 3(x-1)^2 + 4(y+1)^2 = 12</p>
<p>This is an equation for an ellipse centered at (1,-1), with a horizontal 'long' axis. </p>
<p>( x - 2 )^2 = 4( y + 2 ) represents a parabola with a min at (2,-2). Plug in (2,-2) into the LHS of the ellipse equation. If the LHS value is < 12, (2,-2) lies within the ellipse; if it's > 12, it lies outside the ellipse. You actually get LHS=7, which is enough to show that (2,-2) lies within the ellipse.</p>
<p>You will therefore have 2 points of intersection (one where the parabola enters the ellipse, one where it leaves).</p>
<p>well the number of points at which the ellipse and parabola cross each other can be 4 too coz see the parabola is vertically placed and the ellipse's upper and lower points can cross the parabola.. then?</p>
<p>Sure - but then the vertex of the parabola would lie outside the ellipse. If that happened, you could have 0 intersection points (the parabola doesn't intersect the ellipse at all) or 2 (the parabola enters & exits the ellipse once, reaches its vertex outside the ellipse, and misses the ellipse on the way back) or 4 (the parabola enters & exits the ellipse once, reaches its vertex outside the ellipse, and enters & exists the ellipse on the way back as well).</p>
<p>Actually, you could even have 1 (the parabola 'touches' the ellipse once only) or 3 points (the parabola enters/exits once, & touches the ellipse once on the way back). Oy, what a headache...</p>
<p>got my test on oct 8.. got an easier way out???</p>
<p>Not for this particular problem. The quickest way that I can see is to check if the parabola vertex lies within the ellipse. If yes, the answer is easy; if not, I don't see any easy way out.</p>
<p>...actually, you kids have it easy these days, with advanced technology like graphing calculators. Back in my day, we had to chip out the curves on stone surfaces :) .</p>
<p>makes sense 2 buy a grpahic calci rite now or should i forget about such some which are usually rare</p>
<p>btw thnx for d help fred flintstones..</p>
<p>Are students permitted to use a scientific calculator in the Math II test in which they have entered--before coming into the test site--the quadratic equation, for example, and perhaps other possibly useful equations?</p>
<p>If you have a link to the collegeboard rule, please post it. If there is no collegeboard rule on this, how are test takers supposed to know what's permitted and what is not? And how does the collegeboard ensure that no students have either an advantage or a handicap?</p>
<p>TIA.</p>
<p>scientific and graphic calcis are allowed for the test.. go to the site and see under ths materials to bring for the test</p>
<p>I think TI-89 should be banned on the SAT, but since it's allowed so far, here how it goes:</p>
<p>solve(3x^2 + 4y^2 -6x + 8y - 5 = 0 and ( x - 2 )^2 = 4( y + 2 ), x).</p>
<p>7 sec. later:</p>
<p>x=-(sqrt(3)-1) and y=(sqrt(3)-2)/2 or
x=sqrt(3)+1 and y=-(sqrt(3)+2)/2.</p>
<p>That's it... Two points.</p>
<p>Pretty sad.</p>
<p>Looks like you might want to work on reading comprehension.</p>
<p>dude, you'd first need to solve for y in each equation and then plug them into your calc.</p>
<p>prachi240987 -
Namaste.</p>
<p>You should not worry that such a labor intensive question will show up on the real test. Solving it for practive does help master a bunch of concepts (ellipse equation, parabola vertex, graphs intersecion, etc.)
As far as graphing calculator, it's quite helpful on the SAT MATH 2, bu it would probably make sense to buy it for the next take of the MATH 2 if you are not happy with your score this time.</p>
<p>Wish you to do well this time!</p>
<p>thnx all u guys for your help</p>
<p>what labor intensive ....lemme tell the quickest way to solve this prob......</p>
<p>The vertex of parabola is (2,-2) .... put that value in the equaion of the circle and find the value:</p>
<p>If value =x
and x<0 then point is within the circle
if x>0 then point is outside the circle</p>
<p>Here x= -5 <0 so within the circle......</p>
<p>Well for a parabola with vertex within a circle.... how many possible intersection points can be there......Maximum two</p>
<p>Do u even need a calculator for this</p>
<p>smartmind -
I stand corrected.
I guess I was thinking of the worst case scenarion: the vertex of parabola is outside the ellipse.</p>
<p>Plugging (2,-2) in the given formula for ellipse
3x^2 + 4y^2 -6x + 8y - 5 =
3(2)^2 + 4(-2)^2 - 6(2) + 8(-2) - 5 = -37 < 0.
The vertex is inside the ellipse, thus two points of intersection.</p>
<p>Optimizerdad used the same fact:
point (x1,y1) lies inside the graph of f(x,y)=0
if f(x1,y1)<0. </p>
<p>He also showed a useful technique of completing the square to get the equation to the standard form. It does help to know that, and also how to interprete the standard equation for the conics (finding center, foci, etc.).</p>
<p>Similar example:
how many common points do these graphs have:
|x| + |y| = 4 and
y + 3 = 3(x + 2)^2?</p>