Equation of a line which is tangent to both the curve `y=x^2+1\ a n d\ y=x^2` is `y=sqrt(2)x+1/2` (b) `y=sqrt(2)x-1/2` `y=-sqrt(2)x+1/2` (d) `y=-sqrt(2)x-1/2`

Equation of a line which is tangent to both the curve y=x^(2)+1 and y=x^(2) is y=sqrt(2)x+(1)/(2) (b) y=sqrt(2)x-(1)/(2)y=-sqrt(2)x+(1)/(2)(d)y=-sqrt(2)x-(1)/(2)

The equation of tangent to the curve y=int_(x^(2))^(x^(3))(dt)/(sqrt(1+t^(2))) at x=1 is sqrt(3)x+1=y (b) sqrt(2)y+1=x sqrt(3)x+y=1(d)sqrt(2)y=x

The length of the tangent to the curve y=f(x) is equal to: A) ysqrt(1+(y1)^2) B) (y/(y1))sqrt(1+(y1)^2) C) ((y1)/y)sqrt(1+(y1)^2) D) y1sqrt(1+(y1)^2)

The equation of a circle of radius 1 touching the circles x^2+y^2-2|x|=0 is (a) x^2+y^2+2sqrt(2)x+1=0 (b) x^2+y^2-2sqrt(3)y+2=0 (c) x^2+y^2+2sqrt(3)y+2=0 (d) x^2+y^2-2sqrt(2)+1=0

The equation of a circle of radius 1 touching the circles x^2+y^2-2|x|=0 is (a) x^2+y^2+2sqrt(2)x+1=0 (b) x^2+y^2-2sqrt(3)y+2=0 (c) x^2+y^2+2sqrt(3)y+2=0 (d) x^2+y^2-2sqrt(2)+1=0

The equation of a circle of radius 1 touching the circles x^2+y^2-2|x|=0 is (a) x^2+y^2+2sqrt(2)x+1=0 (b) x^2+y^2-2sqrt(3)y+2=0 (c) x^2+y^2+2sqrt(3)y+2=0 (d) x^2+y^2-2sqrt(2)+1=0

A circle C of radius 1 is inscribed in an equilateral triangle PQR . The points of contact of C with the sides PQ, QR, RP and D, E, F respectively. The line PQ is given by the equation sqrt(3) +y-6=0 and the point D is ((sqrt(3))/(2), 3/2) . Equation of the sides QR, RP are : (A) y=2/sqrt(3) x + 1, y = 2/sqrt(3) x -1 (B) y= 1/sqrt(3) x, y=0 (C) y= sqrt(3)/2 x + 1, y = sqrt(3)/2 x-1 (D) y=sqrt(3)x, y=0

if x=sqrt(3)+(1)/(sqrt(3)) and y=sqrt(3)-(1)/(sqrt(3)) then x^(2)-y^(2) is

If two different tangents of y^2=4x are the normals to x^2=4b y , then (a) |b|>1/(2sqrt(2)) (b) |b| 1/(sqrt(2)) (d) |b|<1/(sqrt(2))

If two different tangents of y^2=4x are the normals to x^2=4b y , then (a) |b|>1/(2sqrt(2)) (b) |b| 1/(sqrt(2)) (d) |b|<1/(sqrt(2))