For any non-zero real value of `m` ,the equation of the parabola to which the line `mx-y+10+m^(2)=0` is a tangent, is

If the equation m^(2)(x + 1) + m(y – 2) + 1 = 0 represents a family of lines, where ‘m’ is parameter then find the equation of the curve to which these lines will always be tangents.

Statement 1:y=mx-(1)/(m) is always a tangent to the parabola,y^(2)=-4x for all non-zero values of m.Statement 2: Every tangent to the parabola.y^(2)=-4x will meet its axis at a point whose abscissa is non- negative.

The equation of the tangent to the parabola y^(2)=8x which is perpendicular to the line x-3y+8=0 , is

The equation of the tangent to the parabola y^(2) = 8x, which is parallel to the line 2x-y+7=0 ,will be

The area of the region enclosed between parabola y^(2)=x and the line y=mx is (1)/(48) . Then the value of m is

Using integration,find the value of m. If the area bounded by parabola y^(2)=16ax and the line y=mx is (a^(2))/(12) square units.

For all real values of m, the straight line y=mx+sqrt(9m^(2)-4) is a tangent to the curve :

Statement I The line y=mx+a/m is tangent to the parabola y^2=4ax for all values of m. Statement II A straight line y=mx+c intersects the parabola y^2=4ax one point is a tangent line.