SATHEE: Application of Derivatives 4 Question 60

Application of Derivatives 4 Question 60

####63. A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ at the points $P$ and $Q$ . If the tangents to the ellipse at $P$ and $Q$ meet at the point $R$ .

If $Δ (h) =$ area of the $\triangle P Q R, \Delta_{1}=\max {1 / 2 \leq h \leq 1} \Delta(h) $a n d$ \Delta{2}=\min {1 / 2 \leq h \leq 1} \Delta(h) $, t h e n$ \frac{8}{\sqrt{5}} \Delta{1}-8 \Delta_{2}$ is equal to (2013 Adv)

Show Answer

Solution:

PLAN As to maximise or minimise area of triangle, we should find area in terms of parametric coordinates and use second derivative test.

Here, tangent at $P (2 \cos θ, \sqrt{3} \sin θ)$ is

$\begin{aligned} \frac{x}{2} \cos θ + \frac{y}{\sqrt{3}} \sin θ = 1 \\ ∴ R (2 \sec θ, 0) \\ \Rightarrow Δ = Area of △ P Q R \\ = \frac{1}{2} (2 \sqrt{3} \sin θ) (2 \sec θ - 2 \cos θ) \\ = 2 \sqrt{3} \cdot \sin^{3} θ / \cos θ \end{aligned}$

When

$\frac{1}{4} \leq \cos θ \leq \frac{1}{2}$

$∴ Δ_{1} = Δ_{max}$ occurs at $\cos θ = \frac{1}{4} = \frac{2 \sqrt{3} \cdot \sin^{3} θ}{\cos θ}$

When $\cos θ = \frac{1}{4} = \frac{45 \sqrt{5}}{8}$

$Δ_{2} = Δ_{min}$ occurs at $\cos θ = \frac{1}{2}$

$= \frac{2 \sqrt{3} \sin^{3} θ}{\cos θ}$

When $\cos θ = \frac{1}{2} = \frac{9}{2}$

$∴ \frac{8}{\sqrt{5}} Δ_{1} - 8 Δ_{2} = 45 - 36 = 9$

Application of Derivatives 4 Question 60

Solution:

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 60

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu