JEE MATHS TOUR

 Foot of Perpendicular from a given point on a line Let  P : (x1 ,  y1) be the point and L : (ax + by + c = 0) be the line . Let (x , y) be the foot of perpendicular.  Method 1 : Let  '' m1 ''  be the slope of  line '' L ''. m1 = -a / b Now , we make a line perpendicular to  '' L '' passing through ''  P  ''. So, slope of that line is (-1 / m1) and passing point point is (x1 , y1). We can find line equation by using point slope form. And, then find the intersection point of given line and above line. But this method is time consuming. Method 2 :  Example : Let 2x + y = 3 be the line and (1, 2) be the point . So, L :  2x + y - 3 = 0   { a = 2  ;   b = 1 ;  c = -3} P : (1 ,  2) {x1 = 1 ;  y1  = 2} Now,  x - 1  =  y - 2  =  - 1     2           1           5      x - 1 ...

         Finding    common   tangents    b/w    two    circles Q.). What is the meaning of  common tangents ? Ans.)  Its  meaning is as simple as its name, a common tangent is a tangent that is tangent to both the given curves. These curves may be two parabolas , one parabola and a circle or may be hyperbola and a parabola. Actually, there are immense possibility what these curves can be. But we will specifically discuss about two circles in this post. There are maximum " 4 " possible common tangents b/w to circles and minimum is zero. OK lets understand by diagrams. Now, we will discuss how to calculate number of common tangents b/w two circles : Let C1 and C2 be two circles with r1 and r2 be there respective radius and (x1, y1) and (x2, y2) be the centre of C1 and C2 respectively. Let " D " be the distance b/w centre of two circles C1 and C2. Now, If 1)            ...

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Unable to find solution to some difficult Maths problems and has become nightmare(👿) ? No problem because we are here to solve them, just ask and we will provide you most detailed solutions like this :  Q.)A line parallel to the straight line 2x - y = 0 is tangent to hyperbola P at point (x1 , y1). Then x 1 ^2 + 5y 1 ^2 is equal to : a) 6 b) 8 c) 10 d) 5 where P :  Solution:  Let's draw a sketch for hyperbola P and line: Tangent to point (x1, y1) to P is :  x.x1/4 - y.y1/2 = 1 As this is parallel to given line both have same slope. So, x 1 /2y 1 = 2  --->   x 1  = 4y 1                            --------- (1) and also point (x 1 , y 1 ) also satisfy P.  So, x 1 ^2 / 4 - y 1 ^2 / 2 = 1                                --------- (2) From above (1) and (2) : y 1 ^2  =  2/7 As, we ...