If the slope of one of the lines represented byax2+ 2hxy + by2= 0 is the square ofthe other then ab(a+ b) + 8h3=(1) 4 abh(2) 5 abh(3) 6 abh(4) 8 abh71.If 9x2+ 2hxy + 4y2+ 6x + 2fy – 3 = 0represents two parallel lines then(1) h = 6, f = 2(2) h = –6, f = 2(3) h = 6, f = –2(4) None72.Two points A and B move on the x-axis andthe y-axis respectively such that the distancebetween the two points is always the same.The locus of middle point of AB is(1) a straight line(2) a pair of straight lines(3) a circle(4) none of these73.A family of lines is given by (1 + 2O)x +(1 –O)y +O= 0,Obeing the parameter. Theline belonging to this family at the maximumdistance from the point (1, 4) is(1) 4x – y + 1 = 0(2) 33x + 12y + 7 = 0(3) 12x + 33y = 7(4) None of these74.If the points (–2, 0),1,3³§·²¨¸©¹and (cosT, sinT)are collinear then the number of values ofT[0, 2S] is(1) 0(2) 1(3) 2(4) infinite75.The values ofkfor whichlineskx + 2y + 2 = 0, 2x + ky + 3 = 0,3x + 3y + k = 0 are concurrent(1) {2, 3, 5}(2) {2, 3, –5}(3) {3, –5}(4){–5}76.Equation of angle bisector of the lines3x – 4y + 1 = 0 and 12x + 5y – 3 = 0 containingthe point (1, 2) is(1) 3x + 11y – 4 = 0(2) 99x – 27y – 2 = 0(3) 3x + 11y + 4 = 0(4) 99x + 27y – 2 = 077.The co-ordinates of the point of reflection ofthe origin (0, 0) in the line 4x – 2y – 5 = 0 is :(1) (1, –2)(2) (2, –1)(3)42,55§·²¨¸©¹(4) (2, 5)  Text Solution `8h^(2)-9ab``8h^(2)=9ab^(2)``8h=9ab``8h=9ab^(2)` Answer : A Solution : We have, `ax^(2)+2hxy+by^(2)=0` <br> Let slope of one line is m <br> `:.` Slope of another line is 2 m <br> We know that, <br> `m_(1)+m_(2)=-(2h)/(b)` <br> and `m_(1)m_(2)=a/b` <br> `:.m+2m=-(2h)/(b)` <br> `rArr 3m=(-2h)/(b)` ...(i) <br> and `m(2m)=(a)/(b)` <br> `rArr 2m^(2)=a/b` ...(ii) <br> On eliminating m , we get <br> `2((-2h)/(3b))^(2)=a/b` <br> `rArr 8h^(2)=9ab` If the slope of one of the lines given by `ax^2 + 2hxy +by^2 = 0` is two times the other, then 8h2 = 9ab. Explanation: Given equation of pair of lines is `ax^2 + 2hxy +by^2 = 0` ∴ `m_1 + m_2 = (-2h)/b` and m1m2 = `a/b` According to the given condition, `m_1 = 2m_2` ∴ `2m_2 + m_2 = (-2h)/b` and `2m_1m_2 = a/b` ∴ `m_2 = (-2h)/(3b)` and `m_2^2 = a/(2b)` ∴ `((-2h)/(3b))^2 = a/(2b)` ∴ `(4h^2)/(9b^2) = a/(2b)` ∴ 8h2 = 9ab 1) 4λh = ab(1 + λ)
2) λh = ab(1 + λ)2
3) 4λh2 = ab(1 + λ)2
4) None of these
Solution:
Given ax2 + 2hxy + by2 = 0 ..(i) Let m be the slope. Then the other slope is λm. We know sum of slopes, m1 + m2 = -2h/b => m + λm = -2h/b => m(1 + λ) = -2h/b => m = -2h/b(1 + λ) …(ii) Product of slopes, m1m2 = a/b => λm2 = a/b => m2 = a/bλ….(iii) Squaring (ii) and equating to (iii) 4h2/b2(1 + λ)2 = a/bλ => 4λh2= ab(1 + λ)2 Hence option (3) is the answer.
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If the slope of one of the lines represented by ax^2 + 2hxy + by^2 = 0 is two times the other
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