IGDTUW 2026 — Mathematics PYQ

Step 1: Write the general equation of a tangent to the first parabola

The first parabola is:

$$y^2=4x$$

Comparing this with the standard form $y^2=4ax$, we get:

$$4a=4⟹a=1$$

The equation of a tangent to the parabola $y^2=4ax$ in slope form ($m$) is given by:

$$y=mx+\frac{a}{m}$$

Substituting $a=1$:

$$y=mx+\frac{1}{m}\text{— (Equation 1)}$$

Step 2: Apply the condition of tangency to the second parabola

Since this same line touches the second parabola $x^2=-32y$, we can substitute the value of $y$ from Equation 1 into the equation of the second parabola.

Substitute $y=mx+\frac{1}{m}$ into $x^2=-32y$:

$$x^2=-32\left(mx+\frac{1}{m}\right)$$

$$x^2=-32mx-\frac{32}{m}$$

Rearranging it into a standard quadratic equation in terms of $x$:

$$x^2+32mx+\frac{32}{m}=0$$

Step 3: Set the discriminant to zero for tangency

For a line to be tangent to a curve, the quadratic equation formed by their intersection must have equal roots. Therefore, its discriminant ($D=b^2-4ac$) must be equal to zero.

Here:

• $A=1$

• $B=32m$

• $C=\frac{32}{m}$

Setting $D=0$:

$$B^2-4AC=0$$

$$(32m{)}^2-4(1)\left(\frac{32}{m}\right)=0$$

$$1024m^2-\frac{128}{m}=0$$

Multiply the entire equation by $m$ (since $m\ne 0$):

$$1024m^3-128=0$$

$$1024m^3=128$$

$$m^3=\frac{128}{1024}$$

$$m^3=\frac{1}{8}$$

Taking the cube root on both sides:

$$m=\frac{1}{2}$$

Step 4: Find the final equation of the common tangent

Substitute the value of $m=\frac{1}{2}$ back into Equation 1:

$$y=\frac{1}{2}x+\frac{1}{\frac{1}{2}}$$

$$y=\frac{1}{2}x+2$$

Multiply the entire equation by 2 to clear the fraction:

$$2y=x+4$$

Rearranging it into standard general form:

$$x-2y+4=0$$

Correct Answer

The correct option is (d).