JEE Advanced
JEE Main
BITSAT
View complete IIT JEE section
NTSE
KVPY
Olympiads
CBSE
ISCE
UAE
Saudi Arabia
Qatar
Kuwait
Oman
Bahrain
View complete NRI section
Physics
Chemistry
Maths
Revision notes
View complete study material
Buy IIT JEE/AIPMT study material
Buy CBSE Grade 8/9/10 study material
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)...
Ans: 4 Hello student, Please find answer to your question below Equation of ellipse: Equation of normal: Let the normal passes through point (x 1 , y 1 ) This polynomial equation in theta have four...
Ans: Hello student, Please find answer to your question below …...........(1) ............(2) Angle b/w two lines: Let the slope of the lines be ‘m’ which make equal angle with...
Ans: Hello Student, Please find answer to your question below Let the centre of 1 st circle to be (a, b) Then, radius of circle be: a – 4 Circle is passing through (7, 2) & (4, b). Equation...
Ans: Hello Student, Please find the answer to the your question below Circle S passes through the point (2, 3). (1, 2) is the centre of S. Radius of circle:
Equation of the line BC passing through (1,2) with slope ‘m’ is (y-2) = m(x-2) Since ABC is isosceles, angle B = angle C. using Cos(theta) formula for finding angle between the lines AB,BC & AC,BC,...
I think there is some problem with your question since the points lying inside a triangle constitute an enclosed area. So these pointscannot lie on a st. line.
This is an equilateral triangle wiht sides root 32. So, the centroid, circumcentre, orthocentre, incentre will coincide. and will lie inside the triangle.
so.. centroid is dividing Orhtocentre and circumcentre in 2:1 internally use section formula if the circumcentre is (x, y) for X-coordinate: 3 = (4*1 + 2* x)/ 3 ….....(1) 9 =(4*1 + 2* x) 5 = 2...
Solution: As we know the.. orthocentre is the point of intersection of altitudes of the triangle... soo for right angle triangle.. orthocenter is the vertex itself on which right angle is made.....
Solution: As we know the.. orthocentre is the point of intersection of altitudes of the triangle... soo for right angle triangle.. orthocenter is the vertex itself on which right angle is made.. here...
O = orthocentre G = centroid C = circumcentre so orthocentre , centroid and circumcentre belong to a straight line , called Line of Euler . thanks and regards
if two lines are: L1 : a 1 x + b 1 y + c 1 = 0 L2 : a 2 x + b 2 y + c 2 = 0 then eqn of angle bisectors (a1x + b1y + c1) / {(a 1 ) 2 + (b 1 ) 2 } 1/2 = + - (a2x + b2y + c2) / {(a2)^2+ (b2)^2}^1/2 we...
solution: given L1 :24x+7y=20 ; slope M1 = -24/7 L2 ;4x-3y=2 ; slope M2= 4/3 and L3 :2x+11y=5 ; slope M3= -2/11 ifperpendiculars falling from any point from the line L3 to L1 and L2 areequal to each...
solution: as shown.. the equn of the line (x – x 1 ) / cos A = ( y – y 1 ) / sin A = R where A = angle of inclination with X-axis R = distance of the point on the line so ( x – 2)/...
soln: point of intersection of its diagonals is center of the square.... as shown in the image below the center is P (2,2) thanks and regards
solution: ans: “a” and “ c” both AS given circle is touching both the axis and passing through (3, -6) so its clear that circle would be in 4 th quadrant .. so the center also...
Dear student, If we compare this with the standard equation: ax 2 +2hxy+by 2 =0 We get, a =1, h = 1, b = -1 Hence using the formula: we get: Hence, Regards Sumit
GIVEN QUESTION CAN BE SOLVED USING A.M-G.M INEQUALITY FOR PERIMETER TO BE MINIMUM AM +BM SHOULD BE MINIMUM i.e AM=BM OR SIMPLY M LIES ON PERPENDICULAR BISECTOR OF AB AND LINE2x-3y=9 i.e DEDUCE THE EQN...
Dear student, We know that atany pointP having the coordinates (x,y)on the unit circle satisfies the relation Hence, we get that Now we can use the usual parameterization of the circle, i.e....
this question is to be done by done by the method of family of lines. so,by this method-(x+y+1)+lambda(2x-3y-1)=0 now x=--2/5 and y=--3/5 are the solutions of the equation x+y+1 =0 and 2x-3y=1. so,put...
Hello Student, Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty
Hello Student, clearly Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty
Asymptotes are tangents to a hyperbola which never really touch it or mathematically u can say touch it at infinity Asymptotes are very imp from jee pt of view so do it very nicely good luck :)
982