SATHEE: Application of Derivatives 4 Question 36

Application of Derivatives 4 Question 36

####38. A line $L : y = m x + 3$ meets $Y$ -axis at $E (0, 3)$ and the arc of the parabola $y^{2} = 16 x, 0 \leq y \leq 6$ at the point $F (x_{0}, y_{0})$ . The tangent to the parabola at $F (x_{0}, y_{0})$ intersects the $Y$ -axis at $G (0, y_{1})$ . The slope $m$ of the line $L$ is chosen such that the area of the $△ E F G$ has a local maximum

Match List I with List II and select the correct answer using the codes given below the list.

	Column I	Column II
P.	$m =$	1.	$1 / 2$
Q.	Maximum area	2.	4
	of $△ F F G$ is

Codes

	$P$	$Q$	$R$	$S$
(a)	4	1	2	3
(b)	3	4	1	2
(c)	1	3	2	4
(d)	1	3	4	2

Passage Based Problems

Consider the function $f : (- \infty, \infty) \to (- \infty, \infty)$ defined by $f (x) = \frac{x^{2} - a x + 1}{x^{2} + a x + 1}; 0 < a < 2$ .

$(2008, 12 M)$

Show Answer

Answer:

Correct Answer: 38. (a)

Solution:

Here, $y^{2} = 16 x, 0 \leq y \leq 6$

Tangent at $F, y t = x + a t^{2}$

At $x = 0, y = a t = 4 t$

Also, $(4 t^{2}, 8 t)$ satisfy $y = m x + c$ .

$\begin{aligned} \Rightarrow & 8 t & = 4 m t^{2} + 3 \\ \Rightarrow & 4 m t^{2} - 8 t + 3 & = 0 \\ ∴ & Area of Δ & = \frac{1}{2} | \begin{array}{ccc} 0 & 3 & 1 \\ 0 & 4 t & 1 \\ 4 t^{2} & 8 t & 1 \end{array} | = \frac{1}{2} \cdot 4 t^{2} (3 - 4 t) \\ ∴ & = 2 [3 t^{2} - 4 t^{3}] \\ \frac{d A}{d t} & = 2 [6 t - 12 t^{2}] = - 12 t (12 t - 1) \\ \frac{1}{2} + \frac{1}{2} \end{aligned}$

$∴$ Maximum at $t = \frac{1}{2}$ and $4 m t^{2} - 8 t + 3 = 0$

$\begin{aligned} \Rightarrow & m - 4 + 3 & = 0 \\ \Rightarrow & m & = 1 \\ \Rightarrow G (0, 4 t) & \Rightarrow G (0, 2) \\ y_{1} & = 2 \\ x_{0}, y_{0}) = (4 t^{2}, 8 t) & = (1, 4) \\ y_{0} & = 4 \\ Area & = 2 \frac{3}{4} - \frac{1}{2} = \frac{1}{2} \end{aligned}$

Application of Derivatives 4 Question 36

Codes

Passage Based Problems

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Application of Derivatives 4 Question 36

Codes

Passage Based Problems

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu