Let P(4, –4) and Q(9, 6) be two points on the parabola, y2 = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of \Delta PXQ is maximum. Then this maximum area (in sq. units) is :

Parabola PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q31 with Solution

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Correct answer: B

The parabola is $y^2=4x.$ A standard parametric point on it is $X(t)=(t^2,2t).$

Now check the given points:

So the arc $POQ$ means the portion from $t=-2$ to $t=3$ passing through $t=0$ . Hence let

\qquad -2\le t\le 3.$$ --- 2. **Area of triangle $PXQ$** Using coordinates, $$P=(4,-4),\quad X=(t^2,2t),\quad Q=(9,6).$$ Area of triangle is $$\Delta = \frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$ Substitute: $$\Delta(t)=\frac12\left|4(2t-6)+t^2(6+4)+9(-4-2t)\right|.$$ Simplify: $$\Delta(t)=\frac12\left|8t-24+10t^2-36-18t\right|$$ $$=\frac12\left|10t^2-10t-60\right|$$ $$=5\left|t^2-t-6\right|.$$ Factorize: $$t^2-t-6=(t-3)(t+2).$$ Thus $$\Delta(t)=5|(t-3)(t+2)|.$$ For $-2\le t\le 3$, - $t-3\le 0$ - $t+2\ge 0$ So $(t-3)(t+2)\le 0$, hence $$|(t-3)(t+2)|=-(t-3)(t+2).$$ Therefore $$\Delta(t)=5(3-t)(t+2).$$ --- 3. **Maximize the area** Now $$\Delta(t)=5(3-t)(t+2)=5(-t^2+t+6).$$ This is a downward-opening quadratic in $t$, so its maximum occurs at the vertex: $$t=\frac{-b}{2a} = \frac{-1}{2(-1)}=\frac12$$ for the quadratic $-t^2+t+6$. Now compute maximum area: $$\Delta_{\max}=5\left[-\left(\frac12\right)^2+\frac12+6\right]$$ $$=5\left(-\frac14+\frac12+6\right)$$ $$=5\left(\frac14+6\right)$$ $$=5\cdot \frac{25}{4}$$ $$=\frac{125}{4}.$$ --- 4. **Check options** The maximum area is $$\boxed{\frac{125}{4}}.$$ So the correct option is **B**.

View written solutionFree

Correct answer: B

The parabola is $y^2=4x.$ A standard parametric point on it is $X(t)=(t^2,2t).$

Now check the given points:

So the arc $POQ$ means the portion from $t=-2$ to $t=3$ passing through $t=0$ . Hence let

\qquad -2\le t\le 3.$$ --- 2. **Area of triangle $PXQ$** Using coordinates, $$P=(4,-4),\quad X=(t^2,2t),\quad Q=(9,6).$$ Area of triangle is $$\Delta = \frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$ Substitute: $$\Delta(t)=\frac12\left|4(2t-6)+t^2(6+4)+9(-4-2t)\right|.$$ Simplify: $$\Delta(t)=\frac12\left|8t-24+10t^2-36-18t\right|$$ $$=\frac12\left|10t^2-10t-60\right|$$ $$=5\left|t^2-t-6\right|.$$ Factorize: $$t^2-t-6=(t-3)(t+2).$$ Thus $$\Delta(t)=5|(t-3)(t+2)|.$$ For $-2\le t\le 3$, - $t-3\le 0$ - $t+2\ge 0$ So $(t-3)(t+2)\le 0$, hence $$|(t-3)(t+2)|=-(t-3)(t+2).$$ Therefore $$\Delta(t)=5(3-t)(t+2).$$ --- 3. **Maximize the area** Now $$\Delta(t)=5(3-t)(t+2)=5(-t^2+t+6).$$ This is a downward-opening quadratic in $t$, so its maximum occurs at the vertex: $$t=\frac{-b}{2a} = \frac{-1}{2(-1)}=\frac12$$ for the quadratic $-t^2+t+6$. Now compute maximum area: $$\Delta_{\max}=5\left[-\left(\frac12\right)^2+\frac12+6\right]$$ $$=5\left(-\frac14+\frac12+6\right)$$ $$=5\left(\frac14+6\right)$$ $$=5\cdot \frac{25}{4}$$ $$=\frac{125}{4}.$$ --- 4. **Check options** The maximum area is $$\boxed{\frac{125}{4}}.$$ So the correct option is **B**.