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For many of us, we begin our discussion of conic sections in high school with circles and ellipses. We first learn the equation for a circle is of the general form: (x - h)2 + (y - k)2 = r2where (h, k) is the set of coordinates for the center of the circle and r is the radius of the circle. We then move on to ellipses, and find the general equation is[(x - h)2] / a2 + [(y - k)2] / b2 = 1when a > b (horizontally elongated ellipse; switch the positions of the x and y terms for a vertically elongated ellipse). The length of the major axis is therefore 2a and the minor axis 2b. Here is a horizontally-elongated ellipse (image from Wikipedia):How would one answer the following question?What are the lengths of the major and minor axes for the following ellipse?4x2 + 8x + 16y2 - 64y - 13 = 0We would first complete the square by adding a total of 68 to both sides (split up as 4 and 64) to reveal two factorable trinomials: (4x2 + 8x + 4) + (16y2 - 64y + 64) - 13 = 68Factoring out a 4 and 16 from the first and second trinomials respectively, and adding 13 to both sides, this becomes:4(x2 + 2x + 1) + 16(y2 - 4y + 4) = 81Factoring the trinomials 4(x + 1)2 + 16(y - 2)2 = 81and dividing both sides by 81(4/81)(x + 1)2 + (16/81)(y - 2)2 = 1This is nearly in the form requisite for an ellipse. How do we rewrite this to show a2 and b2 in the denominators? The key is to rewrite the equations first as [(x + 1)2 / (81/4)] + [(y - 2)2 / (81/16)] = 1(Note: 4/81 = 1 / (81/4) and 16/81 = 1/ (81/16))which equals: [(x + 1)2 / (9/2)2] + [(y - 2)2 / (9/4)2] = 1 This ellipse, centered at (-1, 2), has major axis length (2a) of 2(9/2) = 9 whereas the minor axis (2b) measures 2(9/4) = 9/2.For an informative website on conic sections, see http://en.wikipedia.org/wiki/Conic_sectionFor online tutoring help, go to Virtual Tutor Center at http://www.virtualtutorcenter.com

By PhDave LLC

Develop a Time Cushion to Answer More Questions

One of the most common ways people miss out on maximum scores on standardized exams is not necessarily from wrong answers. Rather, it is from lots of questions that remain unanswered once time expires! This is especially detrimental on exams where you do not lose points for incorrect responses on the multiple choice (such as AP exams). Take a typical SAT math section, where you must complete 18 questions in 25 minutes. This means you have an average of (25 min x 60 sec) / 18 questions = 83 seconds per questionWe also know that not every question is a question that is difficult for us. Some of these questions we know the answer to immediately! A conservative estimate is that you can answer 20% of these questions in 10 seconds or less. If this is the case, then let's say 4 questions of 18 are answerable in 10 seconds (total time = 40 seconds). This means you now have an average of (24.3 min x 60 sec) / 15 questions = 104 seconds per questionThus, though it may seem a little disorganized, a viable strategy on the first pass through a section is to only answer questions to which you immediately know the answer. This means you should, on the second pass, have more than enough time to make an educated guess on every question. Don't wildly guess, but just remember that you have more time than you think. Make sure you keep this in mind for AP exams, where you are not penalized for incorrect responses.Visit Virtual Tutor Center at http://www.virtualtutorcenter.com for your online tutoring needs!

By PhDave LLC

Analyzing a Dihybrid Cross: One Trait Exhibits Incomplete Dominance

Students who study genetics are familiar with monohybrid crosses, that is, crosses between two parents of (usually) known genotype to produce offspring of various genotype and phenotype. Typically, the genotypic and phenotypic ratios are easily analyzed: In this cross of pea plants, a yellow seed parent (heterozygous, Yy) is crossed with a homozygous recessive (green, yy) parent to produce offspring pea plants of known phenotypic and genotypic ratios, as shown in the Punnett square above. How about a more difficult problem? Predict the percentage of pink-flowered tall plants (F1 generation) that result from the cross between a red-flowered tall (heterozygous) plant and a white-flowered short plant. Assume tall is dominant to short.Since the offspring (F1 generation) have pink flowers, we note red and white flower color exhibit incompletedominance to give pink-flowered offspring. These must have a heterozygous genotype with respect to flower color. Therefore, the cross is TtRR x ttWW. This dihybrid cross is shown below:The tall pink-flowered plants, all of which have the same genotype (TtRW), are highlighted in red. Statistically, 50% of the F1 generation is predicted to be tall and pink based on the parents of the offspring. The rest of the offspring are pink-flowered and short (ttRW).For more on dihybrid crosses, see http://www.youtube.com/watch?v=FJKHC6wX1Hk.Need online tutoring: try Virtual Tutor Center (http://www.virtualtutorcenter.com)

By PhDave LLC

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