# Math 314, Homework Assignment Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.

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1 Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose l nd l re perpendiculr. Thus, the lines intersect in point, which we denote y P 0 nd its coordintes y (x 0, y 0 ). As neither l nor l is verticl, ech line intersects the verticl line with eqution x = x 0 + in points P nd P, respectively. By definition of slope, the coordintes of P re (x 0 +, y 0 + m ) nd those of P re (x 0 +, y 0 + m ). Since l nd l re perpendiculr, the tringle with vertices P 0, P nd P is right tringle, so tht d(p, P ) = d(p 0, P ) + d(p 0, P ), where d(p, Q) represents the distnce etween points P nd Q. Therefore, (m m ) = () + (m ) + () + (m ) = + m + m. Expnding the left-hnd side nd sutrcting m +m from oth sides, we find m m =. Therefore, m m = s climed. Conversely, ssume m m =. We first must prove tht l nd l intersect. Let Q (x, y ) e point on l nd Q (x, y ) e point on l. Then y y = m (x x ) nd y y = m (x x ) re the equtions for l nd l, respectively. These equtions yield the liner system m x + y = y m x, m x + y = y m x which hs solution since A = m m hs determinnt det A = m + m, which is only 0 if m = m in which cse m m 0. As m m = < 0, y our ssumption, the lines l nd l must intersect in point P 0 (x 0, y 0 ). Let P (x 0 +, y 0 + m ) nd P (x 0 +, y 0 + m ) e the points of intersection of l nd l with the verticl line x = x 0 +. Then, the ngle θ etween l nd l t P 0 is determined y the Lw of Cosines to e cos θ = d(p, P ) d(p 0, P ) d(p 0, P ) d(p 0, P )d(p 0, P ) = (m m ) () + (m ) () + (m ) () + (m ) () + (m ) m m = () + (m ) () + (m ) = 0 since m m =. Therefore, cos θ = 0, so θ = π. Thus l nd l re perpendiculr lines.. This prolem concerns rectngulr hyperols. () Prove tht the eqution of the hyperol x y = cn e written in the form xy = c if we tke the symptotes y = ±x s new x- nd y-xes. Therefore, the hyperol my e prmetrized with prmetric equtions x = ct, y = c/t.

2 Mth 4, Homework Assignment Proof. Let the x -xis e the line y = x nd the y -xis the line y = x. Then x = cos(π/4)x + sin(π/4)y = x + y nd y = sin(π/4)x + cos(π/4)y = x + y descrie the originl vriles x nd y in terms of our new xes vriles, x nd y. Hence, the hyperol x y = my e written in terms of x nd y s = = y x + x + (x ) + (x )(y ) + (y ) = x y. y (x ) (x )(y ) + (y ) Therefore, with respect to our new xes, the eqution for the hyperol x y = hs the form x y = c, where c =, s climed. () Let P nd Q e points on the hyperol with prmeters t nd t, respectively. Determine the eqution of the chord P Q. Solution: The coordintes for P re (ct, c/t ) nd those for Q re (ct, c/t ), c c t respectively. Hence, the slope of the chord P Q is m = t =, y finding ct ct t t common denomintor for the numertor nd cncelling the terms t t in the expression. Therefore, the eqution of the chord P Q is y c t = t t (x ct ). Determine the coordintes of the point N where P Q meets the x-xis. Solution: The point N where the line P Q meets the x-xis must hve y-coordinte 0. Therefore, its x-coordinte is found y solving the eqution of the chord P Q for x when we set y = 0. Tht is, we must solve c = (x ct ) for x. Multiplying t t t oth sides y t t yields ct = x ct, so tht x = ct + ct is the x-coordinte for N. Hence, (ct + ct, 0) re the coordintes of the point N where P Q meets the x-xis. Determine the midpoint M of P Q. Solution: The midpoint of P Q hs coordintes ( c ct + ct t, + c ) ( t ct + ct =, ct ) + ct t t using the midpoint formul. Prove tht OM = MN, where O is the origin.

3 Mth 4, Homework Assignment ( ct + ct Proof. Consider the tringle OMN, where O(0, 0) is the origin, M, ct ) + ct t t is the midpoint of the chord P Q, nd N(ct + ct, 0) is the point where the line P Q intersects the x-xis. If we drw the perpendiculr from( M to the side ) ON, it ct + ct intersects the the x-xis t the point R with coordintes, 0, nd thus isects the chord ON. Thus, OR = RN, ngle ORM equls ngle NRM (oth re right ngles, ecuse MR is perpendiculr to ON), nd MR = MR. Therefore, y the SAS Theorem, tringles ORM nd N RM re congruent. Hence, corresponding sides OM nd NM re equl, y definition of congruence.. Let P e point on the ellipse with eqution x + y =, where > > 0, = ( e ), nd 0 < e <. () If P hs coordintes ( cos t, ), determine the eqution of the tngent t P to the ellipse. Solution: The eqution of the tngent line t the point (x 0, y 0 ) to n ellipse with eqution x + y = is x 0x + y 0y =, y Theorem.. of the ook. Thus, the eqution of the tngent line t P ( cos t, ) is = ( cos t)x + ( )y = x cos t + y. () Determine the coordintes of the point T where the tngent in prt () meets the directrix x = /e. Solution: As the x-coordinte of the point T is x = /e, we my find the corresponding y-coordinte y solving the eqution = /e cos t + y for y. Sutrcting e cos t from oth sides nd multiplying y so T hs coordintes ( e, cos t ). e, we find y = (c) Let F e the focus with coordintes (e, 0). Prove tht P F is perpendiculr to T F. ( ) cos t e, Proof. Let us consider this prolem in two cses. First, suppose tht P F is not verticl, in which cse P F nd T F re non-verticl lines (T F cn never e verticl, s the x- coordinte of F is e while the x-coordinte of T is /e nd e < ). Thus, y Prolem, it suffices to prove tht the product of the slopes of the lines P F nd T F is equl to. Now, the slope of P F is m P F = cos t e nd the slope of T F is cos t m T F = e e e = (e cos t) ( e ).

4 Mth 4, Homework Assignment (e cos t) Hence, m P F m T F = (cos t e)( e ) = ( e ) =, since = ( e ). Therefore, P F is perpendiculr to T F if P F is not verticl. Now ssume P F is verticl, so the x-coordinte of P is cos t = e. Thus cos t = e, so the y-coordinte of T is = 0, which is the sme s the y-coordinte of F. Therefore the line T F is horizontl ( y = 0) whenever P F is verticl, nd so T F is perpendiculr to P F in this cse lso. Thus, the lines P F nd T F re lwys perpendiculr. (d) Determine the eqution of the norml t P to the ellipse. Solution: First, suppose the tngent line x cos t + y = is neither horizontl nor verticl. (This mens tht cos t 0.) Expressing this eqution in slope-intercept form, we find y = cos t x = cos t x +. By Prolem, the slope of the norml line is m = cos t =, so the eqution cos t for the norml line hs the form y = x + B for some vlue of B. Sustituting cos t x = cos t nd y = in for x nd y in the eqution for the norml line ove, we find tht B =, so the eqution of the norml line is y = cos t x +. Multiplying oth sides y cos t gives the eqution ( )x ( cos t)y = ( ) cos t for the norml line. We conclude y considering the cses where the tngent line is either horizontl or verticl. First, if the tngent line t P to the ellipse is horizontl, then its slope is 0, which requires cos t = 0. Thus P hs coordintes (0, ) or (0, ) nd the norml line is the verticl line x = 0. Oserve tht this is the sme s our eqution for the norml line ove, s cos t = 0 gives ( )x = 0 so x = 0. If the tngent line is verticl, then = 0, so = 0. This mens P hs coordintes (, 0) or (, 0). In either cse, the norml line is the horizontl line y = 0, which grees with the formul for the norml line given ove when we set = 0. Thus, regrdless of the loction of P on the ellipse nd the slope of the tngent line t P to the ellipse, the eqution of the norml line is ( )x ( cos t)y = ( ) cos t. (e) Determine the coordintes of the point Q where the norml line in prt (d) meets the xis y = 0. Solution: Setting y = 0 in the eqution of the tngent line found in prt (d) ove, we hve ( )x = ( ) cos t, so tht x = cos t. Thus the point Q hs coordintes ( ) cos t, 0. (f) Let F e the focus with coordintes (e, 0). Prove tht QF = ep F. 4

5 Mth 4, Homework Assignment Proof. Since oth Q nd F re on the x-xis, the length of the segment QF is the solute vlue of the difference of their x-coordintes, The distnce from P to F is QF = e cos t = e ( e ) cos t = e e cos t = e e cos t. P F = ( cos t e) + ( ) = cos t e cos t + e + sin t = cos t e cos t + e + ( e ) sin t = e cos t + e ( sin t) which we see is equl to e QF. = ( e cos t) = e cos t, 4. Let F denote the fmily of prols {(x, y) : y = 4(x+)} s tkes on ll positive vlues, nd G denote the fmily of prols {(x, y) : y = 4( x + )} s tkes on ll positive vlues. Use the reflection property of the prol to prove tht, if F F nd G G, then, t ech point of intersection, F nd G cross t right ngles. Proof. Let F F e the prol {(x, y) : y = 4(x + )} nd G G e the prol {(x, y) : y = 4( x + )}. (Oserve tht the positive numers nd re prmeters tht need not e the sme for F nd G.) Then F is the trnsltion of the prol {(x, y) : y = 4x} y the vector = (, 0) nd G is the trnsltion of the prol {(x, y) : y = 4x} y the vector = (, 0). Hence, the focus of F nd the focus of G coincide, nd oth re the origin, O = (0, 0) R. Suppose P is point of intersection of F nd G. Let l F denote the tngent line to F t P nd l G denote the tngent line to G t P. Let S e the point where the tngent line l F intersects the x-xis nd let T e the point where the line l G intersects the x-xis. By the proof of the Reflection Property of the Prol, the ngles OSP nd OP S re equl nd the ngles OT P nd OP T re equl. Now consider the tringle SP T. The ngle SP T is the sum of the ngles SP O nd OP T, while P ST = P S0 = OSP nd P T S = P T O = OT P. Hence, s the sum of the ngles in tringle is 80, we hve 80 = P ST + P T S + SP T = P SO + P T O + (OP S + OP T ) = (OP S + OP T ), so OP S + OP T = 90. Yet, OP S + OP T is the ngle etween l F nd l G, so l F nd l G re perpendiculr. Therefore, F nd G cross t right ngles, s climed. 5. Clssify the conics in R with the following equtions. Determine the center/vertex nd xis of ech. 5

6 Mth 4, Homework Assignment () x xy + y + 4x 5y + = 0 Solution: By Theorem.., since B 4AC = ( ) 4()() = 5 > 0, this conic is hyperol. To find the center nd xes of the hyperol, we construct the mtrix A =.5.5 nd serch for the specil orthogonl mtrix P tht digonlizes A. This is done y finding the eigenvlues nd eigenvectors of A, so consider the chrcteristic polynomil p A (t) = det(ti A) = (t )(t ) (.5)(.5) = t t.5. Setting p A (t) = 0 nd solving for t, we find tht the eigenvlues of A re λ =.5 nd µ = 0.5. The corresponding eigenvectors re u λ = nd vµ =. Hence P = = so tht P T = P nd det P = s required for P to e specil orthogonl mtrix. Then we perform the chnge of vriles x = x y y = x + y x + y so x = (x + y ) nd y = ( x + y ). Thus, the eqution for the conic in terms of x nd y ecomes 5 (x ) (y ) +4 (x +y ) 5 ( x +y )+ = 5 (x ) (y ) + 9 x y + = 0. Completing the squres, we otin 5 (x ) (y ) + = 9 5, so 5 8 (x ) (y ) + = is the stndrd form of the conic. Hence the center is t x = 9 0 nd y = nd the xes re x = 9 0 (mjor xis) nd y = (minor xis). Chnging vriles ck gin, we see tht x = (x y) nd y = (x+y), so the center of the hyperol is x = 7 5 nd y = 5 nd the xes re x y = 9 5 (mjor) nd x + y = (minor). () x + xy + 4y 7 = 0 Solution: Using Theorem.., we find B 4AC = () 4()(4) = 7 < 0, so this conic is n ellipse. The mtrix A = hs eigenvlues λ = 5+ nd µ = with corresponding eigenvectors u λ = nd v µ = +. Thus P = nd y chnge of vriles x = Px, the eqution of the conic 5 ecomes (x ) + 5 (y ) = 7, (x ) + 5 (y ) =, 4 so the center is x = 0, y = 0 nd the xes re x = 0 (minor) nd y = 0 (mjor). Rewriting this in terms of our originl vriles x = x ( + )y nd y = x+( + )y, the center of the ellipse is (0, 0) nd the xes re x (+ )y = 0 (minor) nd x + ( + )y = 0 (mjor). 6

7 Mth 4, Homework Assignment (c) x + xy + 4y + x 9 = 0 Solution: By Theorem.., s B 4AC = () 4()(4) = 5 < 0, the conic is n ellipse. Then A = , whose eigenvlues re λ = 5 0 nd µ = 5+ 0 with corresponding eigenvectors u λ = nd v µ =. Hence, P =, so x = 0 (x + y ) nd y = ( 0)x + ( + 0)y. 0 Therefore, the eqution of this conic ecomes 5 0 (x ) (y ) + 0(5 0) 0(5 + 0) = C for suitle positive constnt C. Hence the center of this ellipse is t x = 0(5 0), y = 0(5+ nd the xes re 0) x = 0(5 (mjor) nd 0) y = (minor). Putting everything ck in terms of x nd y, the center is t ( 0(5+ 0) , 5 ) while the xes re x + ( 0)y = 5 (mjor) nd x + ( + 0)y = 0 (minor) (d) x + xy + y 7x + = 0 Solution: Using Theorem.., the conic is prol since B 4AC = () 4()() = 0. For this conic, A =, so the eigenvlues re λ = 0 nd µ = nd P =, so tht the eqution ecomes (y ) 7 (x + y ) + = 0. Therefore, y completing the squre, we otin the eqution (y ) 7 4 = 7 (x ) = 7 (x ). Hence, the vertex of the prol is t x = 90 4, y = 7 4 nd the xis of the prol is y = 7 4. Putting everything ck in terms of x nd y, we hve tht the vertex of the prol is t x =.95, y = 0.05 nd the xis of the prol is x + y = 7 4. (e) x xy y = 0 Solution: Since B 4AC = ( ) 4()( ) = 7 > 0, the conic is hyperol y Theorem... Then A = , so tht its eigenvlues re λ = 7 nd µ = 7 nd P = Thus the eqution of the conic, in terms of x nd y is 7 (x ) 7 (y ) =, so 7 4 (x ) 7 4 (y ) =. Thus the center is t x = 0, y = 0 nd the mjor xis is x = 0 while the minor xis is y = 0. Putting things ck in terms of x nd y, we hve tht the center is t x = 0, y = 0 nd the xes re x+( 7 4)y = 0 (mjor) nd x + ( 7 + 4)y = 0 (minor). 7

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