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Sir , can you explain this answer with more steps.
Definition says that the asymptotes of rectangluar hyperbolar are cordinate axis i.e x and y axis By the definition of asymptotes hyperbola meets it at infinity. So rectangular hyperbola intersects...
Shortest focal chord is chord perpendicular to axis and passing through focus. for above parabola focus is (8,0) End points of focus will be (8,16) and (8, -16) ( Get y by putting x=8 iin parabola...
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Hello Student, Please find the solution Here you can use the concept of family of lines The equation of the line passing through the inersection of lines x-y+1=0 and 2x-3y+5=0 can be assumed as Now...
Let us first solve for end points of the chord by finding points of intersection of chord and parabola: It is also interesting to note Let’s now call the slope of tangents at these two points...
A few results will help solving this problem Here, a,b,c are sides of the triangle. s is the semi-radius and r is the in-radius. Using these equations and given condition, we get 3s=4c which gives...
Hello Student, Please find the proof of this: Let us draw (construct) the straight line GH passing through thepoint C parallel to the triangle side AB. Such a line does exist due to postulate 1 of the...
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Let y=mx + c be the equation of chord Homogenize the equation of curve 3x 2 -y 2 -2x+4y=0 by putting 1= (y-mx)/c Equation becomes 3x 2 -y 2 -2x(y-mx)/c+4y(y-mx)/c=0 Homogenization gives the equation...
A (1,5) B (2,3) C (4,4) AB= √5 units BC=√5 units AC= √10 units Hence given triangle is right angled triangle ( AB2 + BC2 = AC2) at B Orthocenter is B only CIrcumcenter for rt...
A (1,5) B (2,3) C (4,4) AB= √5 units BC=√5 units AC= √10 units Hence given triangle is right angled triangle ( AB 2 + BC 2 = AC 2 ) at B Orthocenter is B only CIrcumcenter for rt...
Length of the diagonal of square is √ ((1-2) 2 + (-2+3) 2 +(3-5) 2 ) = √6 We aslo know that length of square is √2 times the side. This implies √2a= √6 ( where a is the...
Dear student, The geometric meaning of eccentric is non-concentric, i.e. deviating from the circular form, as in the elliptical (oval) orbits of the planets. Since for the case of circles the...
yes it is present.........
One diagonal is a member of both the family of lines (x+y-1)+k1(2x+3y-2)=0 and (x-y+2)+k2(2x-3y+5) Hence it must pass through the point of intersection of x+y-1= 0 and 2x+3y-2 = 0 . Also it should...
Equation of focal chord inclined at angle titha with x axis is y = tantitha(x-ae) Let it intersect x^2/a^2 + y^2/b^2 = 1 at P(x1.y1) & Q(x2,y2). Solving we get b^2*x^2 + a^2*tan^2titha*(x-ae)^2 =...
a parallelepiped is a three-dimensional figure formed by six parallelograms. The rectangular cuboid (six rectangular faces), cube (six square faces), and the rhombohedron (six rhombus faces) are all...
Hi, The equation of the normal to the parabola y 2 = 4ax is y = mx – 2am – am 3 , where m is the slope of the tangent line corresponding to the normal. If the normal passes through the...
Pl. check this problem. The circumcentre for the given triangle cannot be (0,0). Also, for finding orthocentre, circumcentre is not required.
see, if (a,b)are the x n y axes,a 2 +b 2 =l 2 .....(c,d) midpt....c=a/2,,d=a/2....c^2+d^2=l^2/4....x^2+y^2=l 2 /4
y 2 = px y = sqrt(px) Since x & y are perfect square integers, sqrt(x) is an integer. Since p is a prime no., sqrt(p) is an irrational no. The point which satisfies the given condition is only (0, 0)....
Sum of the ordinates of conormal points is zero. we can prove by following approach: Equation of normal at any point(2t^2,4t) to the parabola y^2=8x is y= -ty+4t+2t^3 it passes through (3,5) 5=...
*find the angle??a correction in the question ..
a petty question on homogenization of equation. ax 2 +2hxy+by 2 +2gx(1)=0 ...1 a1x 2 +2h1xy+b1y 2 = -2g1x ...2 a1x 2 +2h1xy+b1y 2 /(-2g1x)=1 ...3 now replacing 1 in eq 1 by a1x 2 +2h1xy+b1y 2 /(-2g1x)...
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