Question

What will be the equation of that chord of ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{9} = 1$which passes from the point $(2,1)$ and bisected on the point

• A. $x + y = 2$
• B. $x + y = 3$
• C. $x + 2y = 1$
• D. $x + 2y = 4$

Answer 1

Answer: (D)

$x + 2y = 4$

Answer 2

Let required chord meets to ellipse on the points P and Q whose coordinates are $(x_{1},y_{1})$ and $(x_{2},y_{2})$ respectively

$\because$ Point (2,1) is mid point of chord PQ

∴ $2 = \frac{1}{2}(x_{1} + x_{2})$ or $x_{1} + x_{2} = 4$and$1 = \frac{1}{2}(y_{1} + y_{2})$ or

$y_{1} + y_{2} = 2$

Again points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are situated on ellipse; ∴$\frac{x_{1}^{2}}{36} + \frac{y_{1}^{2}}{9} = 1$and $\frac{x_{2}^{2}}{36} + \frac{y_{2}^{2}}{9} = 1$

On subtracting $\frac{x_{2}^{2} - x_{1}^{2}}{36} + \frac{y_{2}^{2} - y_{1}^{2}}{9} = 0$ or

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = - \frac{(x_{2} + x_{1})}{4(y_{2} + y_{1})} = \frac{- 4}{4 \times 2} = \frac{- 1}{2}$

∴ Gradient of chord $PQ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 1}{2}$

Therefore, required equation of chord $PQ$is as follows, $y - 1 = - \frac{1}{2}(x - 2)$ or $x + 2y = 4$

Alternative: $S_{1} = T$ (If mid point of chord is known)

∴ $\frac{2^{2}}{36} + \frac{1^{2}}{9} - 1 = \frac{2x}{36} + \frac{1y}{9} - 1$ ⇒ $x + 2y = 4$