Find the equation of a common tangent to the parabola y²=2x and the circle x²+y²+4x=0....

eq of tangent to parabola is y=mx+a/m

here a=1/2

=> mx-y+1/2m=0

this is a tangent to the circle.

so , 2=|-2m+1/2m| /

find value of 'm' from above eq. put it in mx-y+1/2m=0 u'll get the eq of common tangent.

Two vertices of a triangle are (5, -1) and (-2, 3). If the orthocentre of the triangle is at origin then the third vertex....

the third vertex is the orthocentre of the triangle formed by the points (5, -1), (-2, 3) and origin.

if in the triangle ABC ,the incentre is the middle point of the median AD, then cosA has the value

incentre divides the median AD in the ratio (b+c):a

its given tht incentre is the midpoint of AD hence a=(b+c)

consider cosA= (b²+c²-a²)/2bc

substitute a=b+c

so cosA= -1

but i dont know how cosA can b "-1" if A is an angle in a triangle.

169{(x-1)² + (y-3)² } = (5x-12y+17)²

it is of d form (SP)²=(PM)²  where S is the focus of the parabola , P is a point on d parabola and PM is perpendicular drawn from P to the directrix

so 5x-12y+17=0 is the eq. of directrix.

and focus is (1,3)

the distance frm focus to directrix is half the length of latus rectum.which is 14/13 

hence length of latus rectum is 28/13

consider one side as base. the opposite side will b the top.

the above two sides cam b painted in ⁶p₂ ways.

4 sides remain..... they can b painted in only 1 way.as cube is symmetric.

so answer is 30.

[1-x^1/2]^1/2/[1+x^1/2]^1/2   dx

= [1-]/

take  = t

= (1-t)(2t)/ dt= 2t/ - t²/ dt

=2/(-2)-2t/ dt + 2 {(1-t²)/  -  1/}dt

=-2 + 2  - tan^-1(t)

= -2 + 2[t/2  + 1/2 sin^-1(t) - tan^-1(t)]

substitute t=