SATHEE: Theory of Equations 4 Question 1

Theory of Equations 4 Question 1

1. Let $P (4, - 4)$ and $Q (9, 6)$ be two points on the parabola, $y^{2} = 4 x$ 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 $△ P X Q$ is maximum. Then, this maximum area (in sq units) is

(2019 Main, 12 Jan I)

(a) $\frac{125}{2}$

(b) $\frac{75}{2}$

(d) $\frac{125}{4}$

Show Answer

Solution:

Given parabola is $y^{2} = 4 x$ ,

Since, $X$ lies on the parabola, so let the coordinates of $X$ be $(t^{2}, 2 t)$ . Thus, the coordinates of the vertices of the triangle $P X Q$ are $P (4, - 4), X (t^{2}, 2 t)$ and $Q (9, 6)$ .

$∴$ Area of $△ P X Q = \frac{1}{2} | | \begin{array}{ccc} 4 & - 4 & 1 t^{2} & 2 t & 1 9 & 6 & 1 \end{array} | |$

$\begin{aligned} = \frac{1}{2} ∣ [4 (2 t - 6) + 4 (t^{2} - 9) + 1 (6 t^{2} - 18 t] ∣ \\ = \frac{1}{2} | [8 t - 24 + 4 t^{2} - 36 + 6 t^{2} - 18 t] | \\ = | 5 t^{2} - 5 t - 30 | = | 5 (t + 2) (t - 3) | \end{aligned}$

Now, as $X$ is any point on the arc $P O Q$ of the parabola, therefore ordinate of point $X, 2 t \in (- 4, 6) \Rightarrow t \in (- 2, 3)$ .

$∴$ Area of $△ P X Q = - 5 (t + 2) (t - 3) = - 5 t^{2} + 5 t + 30$

$[∵ | x - a | = - (x - a), if x < a]$

The maximum area (in square units)

$= - \frac{25 - 4 (- 5) (30)}{4 (- 5)} = \frac{125}{4}$

$[∵$ Maximum value of quadratic expression

$a x^{2} + b x + c, when a < 0 is - \frac{D}{4 a}]$

पिछला

अगला

Theory of Equations 4 Question 1

1. Let $P (4, - 4)$ and $Q (9, 6)$ be two points on the parabola, $y^{2} = 4 x$ 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 $△ P X Q$ is maximum. Then, this maximum area (in sq units) is

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Theory of Equations 4 Question 1

1. 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 △PXQ is maximum. Then, this maximum area (in sq units) is

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

1. Let $P (4, - 4)$ and $Q (9, 6)$ be two points on the parabola, $y^{2} = 4 x$ 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 $△ P X Q$ is maximum. Then, this maximum area (in sq units) is