JEE Main 2021 — Mathematics Coordinate Geometry

$$r=\sqrt{\frac{{p}^{2}}{4}+\frac{{(1-p)}^{2}}{4}-5}$$r=4p2​+4(1−p)2​−5​

$$=\frac{\sqrt{2{p}^{2}-2p-19}}{2}$$=22p2−2p−19​​

Since $r\in (0,5]$r∈(0,5]

So, $0<\frac{\sqrt{2{p}^{2}-2p-19}}{2}\leq 5$0<22p2−2p−19​​≤5

$$0<2{p}^{2}-2p-19\leq 100$$0<2p2−2p−19≤100

$$\Rightarrow 2{p}^{2}-2p-19>0...(i)$$⇒2p2−2p−19>0...(i)

For, $2{p}^{2}-2p-19>0$2p2−2p−19>0

$$p=\frac{2\pm \sqrt{4-4\times 2\times (-19)}}{4}$$p=42±4−4×2×(−19)​​

$$=\frac{2\pm \sqrt{156}}{4}$$=42±156​​

$$\Rightarrow (p-\frac{1+\sqrt{39}}{2})(p-\frac{1-\sqrt{39}}{2})>0....(ii)$$⇒(p−21+39​​)(p−21−39​​)>0....(ii)

$${p}^{2}=10\pm \frac{\sqrt{39}}{2}$$p2=10±239​​

Also, $2{p}^{2}-2p-119\leq 0...(iii)$2p2−2p−119≤0...(iii)

For, $2{p}^{2}-2p-119=0$2p2−2p−119=0

$$p=\frac{2\pm \sqrt{4-4\times 2\times (-119)}}{4}$$p=42±4−4×2×(−119)​​

$$=\frac{1\pm \sqrt{239}}{2}$$=21±239​​

$$\Rightarrow (p-\frac{1+\sqrt{239}}{2})(p-\frac{1-\sqrt{239}}{2})\leq 0...(iv)$$⇒(p−21+239​​)(p−21−239​​)≤0...(iv)

$${p}^{2}=60\pm \frac{\sqrt{239}}{2}$$p2=60±2239​​

So, from (ii)&(iv)

$$\Rightarrow p\in [(\frac{1-\sqrt{239}}{2},\frac{1-\sqrt{39}}{2})\cup (\frac{1+\sqrt{39}}{2},\frac{1+\sqrt{239}}{2})]$$⇒p∈[(21−239​​,21−39​​)∪(21+39​​,21+239​​)]

$$\Rightarrow {p}^{2}\in [(10-\frac{\sqrt{39}}{2},60-\frac{\sqrt{239}}{2})\cup (10+\frac{\sqrt{39}}{2},60+\frac{\sqrt{239}}{2})]$$⇒p2∈[(10−239​​,60−2239​​)∪(10+239​​,60+2239​​)]

So, number of integral values of ${p}^{2}$p2 is $61$61.