CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If the area of the triangle formed by the axes of co-ordinates and tangent at any point P(x, y) in 1st quadrant on the ellipse $$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1$$ be minimum, then find the co-ordinates of P.


A
(2,32)
loader
B
(2,3).
loader
C
(2,3).
loader
D
(3,2).
loader

Solution

The correct option is B $$\left ( 2, 3 \right ).$$
Given equation of ellipse $$\displaystyle \frac{x^{2}}{8}+\frac{y^{2}}{18}=1$$
Here, $$ a^{2}=8, b^{2}=18$$
Let $$P( a\cos t, b\sin t)$$ be a point on the ellipse 
Equation of tangent at $$P( a\cos t, b\sin t)$$ is 
$$\displaystyle \frac{x}{a}\cos t+\frac{y}{b}\sin t= 1$$
Its intercepts on the axes are $$\displaystyle \frac{a}{\cos t}$$ and $$\displaystyle \frac{b}{\sin t}$$
$$\displaystyle \therefore Area\: of \Delta =\frac{1}{2}\frac{ab}{\sin t\cos t}=\frac{ab}{\sin 2t}$$
Area will be least if $$\sin 2t$$ is maximum i.e. $$\displaystyle 2t=90^{\circ}$$ or $$\displaystyle t= 45^{\circ}$$ or $$\displaystyle t= 135^{\circ}$$ (rejected ) as the point P lies in 1st quadrant.
$$\displaystyle \therefore P\: is \left ( \frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}} \right )= \left ( \frac{2\sqrt{2}}{\sqrt{2}}, \frac{3\sqrt{2}}{\sqrt{2}} \right )or \left ( 2, 3 \right ).$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image