IPMAT Indore 2026 · Quantitative Ability · Question 26

Step 1 — find where the two parabolas meet. From {x}^{2}=4ay we get y=\frac{{x}^{2}}{4a}. Substitute into {y}^{2}=4ax:

{\left( \frac{{x}^{2}}{4a}\right)}^{2}=4ax \Rightarrow \frac{{x}^{4}}{16{a}^{2}}=4ax \Rightarrow{x}^{4}=64{a}^{3}x

x\left({x}^{3}-64{a}^{3}\right)=0 \Rightarrow x=0 \text{or}x=4a

Step 2 — discard the origin. The circle is centred at the origin with non-zero radius, so it cannot pass through \left( 0,0 \right). Take x=4a:

y=\frac{{\left( 4a \right)}^{2}}{4a}=4a \Rightarrow \text{the point is}\left( 4a,\,4a \right)

Step 3 — find the radius.

{r}^{2}={\left( 4a \right)}^{2}+{\left( 4a \right)}^{2}=32{a}^{2}

{x}^{2}+{y}^{2}=32{a}^{2}

Answer: Option 4 — {x}^{2}+{y}^{2}=32{a}^{2}

Shortcut. These two parabolas are reflections of each other in y=x, so their non-origin intersection must lie on y=x. Set y=x in either equation: {x}^{2}=4ax gives x=4a at once, and then {r}^{2}=2{\left( 4a \right)}^{2}. The “non-zero radius” clause exists purely to rule out the origin.