The point of intersection of tangent to the curve `y=x^4` at `(1,1)` with the normal to the same curve at `(-1,1)` is (A) `((17)/(15),(13)/(15))` (B) `(7/5,3/5)` (C) `((17)/(15),(23)/(15))` (D) does not exist

The point of intersection of tangent to the curve y=x^(4) at (1,1) with the normal to the same curve at (-1,1) is (A)((17)/(15),(13)/(15)) ( textrmB ) ((7)/(5),(3)/(5)) ( textrmC ) ((17)/(15),(23)/(15)) (D) does not exist

Find the equation of tangent & normal to the curve y = 5x^4 at (1,5)

Find the sldpe of the tangent to the curve y=x^4-4x^2+8 at the point (1,5)

The point on the curve y=5 + x-x^(2) at which the normal makes equal intercepts is a) (1,5) b) (0,-1) c) (-1,3) d) (0,5)

(17)/(15)xx(17)/(15)+(2)/(15)xx(2)/(15)-(17)/(15)xx(4)/(15) is equal to (a)0(b)1(c)10(d)11

Area bounded by the curve y=x^3 , the x-axis and the ordinates x" "=" "" "2 and x" "=" "1 is (A) -9 (B) (-15)/4 (C) (15)/4 (D) (17)/4

Area bounded by the curve y=x^3 , the x - axis and the ordinates x=-2 and x=1 is a) -9 b) -(15)/4 c) (15)/4 d) (17)/4

Tangent to the curve y=x^2+6 at a point (1,7) touches the circle x^2+y^2+16x+12y+c=0 at a point Q , then the coordinates of Q are (A) (-6,-11) (B) (-9,-13) (C) (-10,-15) (D) (-6,-7)

Tangent to the curve y=x^2+6 at a point (1,7) touches the circle x^2+y^2+16x+12y+c=0 at a point Q , then the coordinates of Q are (A) (-6,-11) (B) (-9,-13) (C) (-10,-15) (D) (-6,-7)

The area bounded by the region between curves y=x^(2)+2,x=0,x=3,y=x+1 is s(A)(15)/(2)(B)(15)/(4)(C)(15)/(3)(D)(17)/(2)