SATHEE: Ellipse 2 Question 20

Ellipse 2 Question 20

20. Let $A B C$ be an equilateral triangle inscribed in the circle $x^{2} + y^{2} = a^{2}$ . Suppose perpendiculars from $A, B, C$ to the major axis of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ meets the ellipse respectively at $P, Q, R$ so that $P, Q, R$ lie on the same side of the major axis as $A, B, C$ respectively. Prove that, the normals to the ellipse drawn at the points $P, Q$ and $R$ are concurrent.

$(2000, 7$ M)

Show Answer

Answer:

Correct Answer: 20. (d)

Solution:

Let the coordinates of $A \equiv (a \cos θ, b \sin θ)$ , so that the coordinates of

$B = a \cos (θ + 2 π / 3), a \sin (θ + 2 π / 3)$

and $C = a \cos (θ + 4 π / 3), a \sin (θ + 4 π / 3)$

According to the given condition, coordinates of $P$ are $(a \cos θ b \sin θ)$ and that of $Q$ are $a \cos (θ + 2 π / 3) $, $ b \sin (θ + 2 π / 3)$ and that of $R$ are $a \cos (θ + 4 π / 3), b \sin (θ + 4 π / 3)$

$[∵$ it is given that $P, Q, R$ are on the same side of $X$ -axis as $A, B$ and $C]$

Equation of the normal to the ellipse at $P$ is

$\frac{a x}{\cos θ} - \frac{b y}{\sin θ} = a^{2} - b^{2}$

or $a x \sin θ - b y \cos θ = \frac{1}{2} (a^{2} - b^{2}) \sin 2 θ$

Equation of normal to the ellipse at $Q$ is

$a x \sin θ + \frac{2 π}{3} - b y \cos θ + \frac{2 π}{3}$

$= \frac{1}{2} (a^{2} - b^{2}) \sin 2 θ + \frac{4 π}{3}$

Equation of normal to the ellipse at $R$ is

$a x \sin (θ + 4 π / 3) - b y \cos (θ + 4 π / 3)$

$= \frac{1}{2} (a^{2} - b^{2}) \sin (2 θ + 8 π / 3)$

But $\sin (θ + 4 π / 3) = \sin (2 π + θ - 2 π / 3)$

$= \sin (θ - 2 π / 3)$

and $\cos (θ + 4 π / 3) = \cos (2 π + θ - 2 π / 3)$

$= \cos (θ - 2 π / 3)$

and $\sin (2 θ + 8 π / 3) = \sin (4 π + 2 θ - 4 π / 3)$

$= \sin (2 θ - 4 π / 3)$

Now, Eq. (iii) can be written as $a x \sin (θ - 2 π / 3) - b y \cos (θ - 2 π / 3)$

$= \frac{1}{2} (a^{2} - b^{2}) \sin (2 θ - 4 π / 3)$

448 Ellipse

For the lines (i), (ii) and (iv) to be concurrent, we must have the determinant

$Δ_{1} = | \begin{array}{rc} a \sin θ & - b \cos θ \\ a \sin θ + \frac{2 π}{3} & - b \cos θ + \frac{2 π}{3} \\ a \sin θ - \frac{2 π}{3} & - b \cos θ - \frac{2 π}{3} \\ \frac{1}{2} (a^{2} - b^{2}) \sin 2 θ \\ \frac{1}{2} (a^{2} - b^{2}) \sin (2 θ + 4 π / 3) \\ \frac{1}{2} (a^{2} - b^{2}) \sin (2 θ - 4 π / 3) \end{array} | = 0$