From a point on x+2y-7=0 tangents are drawn to curve `13(x^(2)+y^(2)-2x+1)=(2x+3y-1)^(2)` are perpendicular then the point is
(a)(19,13) (b) (-19,13) (c) (19,-13) (d)(-19,-13)

sqrt((0.361)/(0.00169))=?(1.9)/(13) (b) (19)/(13)(c)(1.9)/(130) (d) (190)/(13)

Determine the points on the curve x^(2)+y^(2)=13 , where the tangents are perpendicular to the line 3x-2y=0 .

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 slope of the tangent to the curve (y-x^(5))^(2)=x(1+x^(2))^(2) at the point (1,3) is.

The value of a for which (13x-1)^(2)+(13y-2)^(2)=(a^(2)-1)(5x+12y-1)^(2) repesents an ellipse is

Find the equations of the tangent and the normal to the curve y = x^(4) - 6x^(3) + 13x^(2) - 10x + 5 at the point (1, 3)

The equation of tangent to the curve y=3x^(2)-x+1 at (1,3) is

The set of values of a for which (13x-1)^(2)+(13y-2)^(2)=a(5x+12y-1)^(2) represents an ellipse is