NIMCET 2026 — Mathematics PYQ

To find the common tangent, let the line equation be $y=mx+c$.

1. Condition for Tangency to $y=-x^2$

A line $y=mx+c$ is tangent to $y=-x^2$ if the quadratic equation $mx+c=-x^2$ (or $x^2+mx+c=0$) has a discriminant equal to zero:

$$D=m^2-4(1)(c)=0⟹c=\frac{m^2}{4}$$

2. Condition for Tangency to $y=(x-2{)}^2$

Expand the parabola equation: $y=x^2-4x+4$.

The line $y=mx+c$ is tangent if $mx+c=x^2-4x+4$ has a discriminant equal to zero:

$$x^2-(4+m)x+(4-c)=0$$

Setting $D=0$:

$$(4+m{)}^2-4(1)(4-c)=0$$

$$16+8m+m^2-16+4c=0⟹m^2+8m+4c=0$$

3. Solving for $m$ and $c$

Substitute $c=\frac{m^2}{4}$ into the second equation:

$$m^2+8m+4\left(\frac{m^2}{4}\right)=0$$

$$m^2+8m+m^2=0⟹2m^2+8m=0$$

$$2m(m+4)=0$$

This gives two possible slopes: $m=0$ or $m=-4$.

• Case 1: If $m=0$, then $c=\frac{0^2}{4}=0$. The tangent is $y=0$.

• Case 2: If $m=-4$, then $c=\frac{(-4{)}^2}{4}=\frac{16}{4}=4$. The tangent is $y=-4x+4$.

Comparing this with the given options, (b) $y=-4x+4$ is the correct equation.