Question

If the equation $\lambda^2 2x + \lambda y - \lambda^2 + 2\lambda + 7 = 0$ represents a family of lines, where '$\lambda$' is parameter, then find the equation of the curve to which these lines will always be tangents.

Answer 1

$(y + 2)^2 - 56x + 28 = 0$

Answer 2

The given equation of the family of lines is: $\lambda^2 2x + \lambda y - \lambda^2 + 2\lambda + 7 = 0$

To find the curve to which these lines are always tangents (i.e., the envelope of the family of lines), we first rearrange the equation into a quadratic form in terms of the parameter $\lambda$: $(2x - 1)\lambda^2 + (y + 2)\lambda + 7 = 0$

This is a quadratic equation in $\lambda$ of the form $A\lambda^2 + B\lambda + C = 0$, where: $A = 2x - 1$ $B = y + 2$ $C = 7$

For the lines to be tangents to a curve, at any point $(x, y)$ on the envelope, there must be exactly one line from the family passing through it. This implies that the quadratic equation in $\lambda$ must have real and equal roots. The condition for equal roots of a quadratic equation is that its discriminant must be zero. Discriminant $D = B^2 - 4AC = 0$

Substitute the expressions for A, B, and C into the discriminant condition: $(y + 2)^2 - 4(2x - 1)(7) = 0$

Now, simplify the equation: $(y + 2)^2 - 56x + 28 = 0$

This is the equation of the curve to which the given family of lines will always be tangents. This equation represents a parabola.