radii vectori, that is, when the pole is at the focus. x+m is the length of the radius vector of a parabola .x or AC = .. x = m + r cos 0 ; .'. r = m + r cos ◊ + m ; By substituting the value of y equal to r sin 0 and x = r cos 0, чу = 0; in the general equation of a circle, viz., y2 +x2 + ay + bx + c the polar equation of the circle will be 2+(a sin + b cos 0) 1+ c = 0. EDALJI NANABHAI. 13. Q. The parabola is an ellipse whose second focus is infinitely distant. A. Let the co-ordinates of the point A be xy, and EC a line parallel to the axis of passing through the point A; then the angle FAC angle EAD; line DF is tangent to parabola; but the angle EAD angle ADB,.. the angle FAC angle ADB. = Take B to be the focus of parabola, and (a y) the co-ordinates of the point A, then AB being a radius vector is equal to x +m; but the distance DG, cut from the axis of x by the tangent DA, is equal to X, therefore DB, the distance of D from the focus, is equal to x+m; but the radius vector BA is equal to x + m .. DB BA, therefore DBA is an isosceles triangle; .. the angle ADB angle DAB; but the angle ADB EAC,.. the angle DAB EAC. We saw in the case of the ellipse, that radii vectori made equal angles with the tangent, and AD and EC in the case of parabola make equal angles with the tangent, therefore EC is parallel to the base DB, and it meets it at infinity, therefore the parabola is an ellipse whose foci are infinitely distant from each other. GOVERDHAN LAXUMAN. 14. Q. Theory of asymptotes? Show that the hyperbola has an asymptote, and that the ellipse and parabola have not. A. Asymptotes are the straight lines that meet the hyperbola at an infinite distance: the equation to the asymptotes is easily obtained : suppose the equation to the straight line is ya x; and that of hyperbola is a2 y2 b2x2 equal to u2 b2 In order to get the intersection of the curve, and the straight line whose equation is yax; substitute the value of y in the equation to the curve, a2 y2 —b2 x2 -a2b2, and we get a2 a2x2 b2 x2 and the value of x out of this equation is - a2 b2; value of a is possible; and again, if it be greater, the value of x is b2 impossible; but if be equal to a2 the value of x is infinite. 2 a b From this it is evident that a must be equal to therefore the a but if we take the equa a2b2, and that to the straight line ya x, the x of the intersection of them is equal to to + 2 a2 + b2 In order to have the equation to the asymptote a2 b2 certain conditions are necessary; a2 + must be equal to no thing; but it is not so, therefore the ellipse has no asymptote. The same reasoning is applicable to the parabola. NARAYAN BALLAL. NORMAL SCHOLARS. Specimens of Answers in Mathematics. 1. Q. Explain the doctrine of limits; give illustrations. sequently, we clearly see, that as h decreases the values of the terms on the other side, commencing with n (n 1) n-2 1.2 creases; and when we make ho, the limit at which we arrive is (n xn−1). VISHWANATH NARAYAN. 2. Q. In the general form of the development of f (x + h), viz. f(x) + A h + Bh2+ &c. show that none of the exponents of h can be fractional. A. For if not, let one of the exponents of h be fractional, then ƒ (x + h) = f (x) + Ah + B h2 + C + &c.: the fourth term has two different values, and one of these plus all the others is = f (x + h) the other value plus the same is also = ƒ (x + h) ; and .. ƒ (x + h) has two different values at the same time, which is absurd. The same may be shown of any other term, and therefore the thing required is done. A. Let z = x.. u= sec z; and differentiating, log z = y log x ; 6. Q. The value of (√ — 1)1 in a series of powers of #? A. We know that ex√-1 π = cos x + sin x ; put x =π e 2 Raise the last equation to 1 power we get e x2 23 π 2 = Now in the series ex- =1+x+ + + &c., put x = 1-2 2 1.2.3 1.2.3 21.2.3.4 24 Ela .&c. 7. Q. Among the values of a which cause the to vanish, to dx determine those which give maximum values of y, and those which give minimum values ? Now in this expression, if the value of x makes f (x) greater than d'y dx2 f(xh), the sign of must be negative, and if the value of x day renders f (x) less than f (x + h), the sign of must be positive. Now, if the same value that makes vanish, also causes to va dx2 nish, then in order that that value may make the function either a d'y maximum or a minimum, it must cause to vanish, and then, as dx3 d'y before, it must make negative for a maximum and positive for a minimum. NARAYAN VISHNU. |