Ellipse
Ellipse
nta_pyq_2025_apr
Grade 11
Question:
The equation of the chord, of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$, whose mid-point is $(3, 1)$ is
$48x + 25y = 169$
$5x + 16y = 31$
$25x + 101y = 176$
$4x + 122y = 134$
Step-by-Step Solution
Key Concept: Apply $T = S_1$: the chord with midpoint $(x_1, y_1)$ on ellipse $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$ is $\tfrac{xx_1}{a^2}+\tfrac{yy_1}{b^2}=\tfrac{x_1^2}{a^2}+\tfrac{y_1^2}{b^2}$.
$T=S_1$: $\dfrac{3x}{25}+\dfrac{y}{16}=\dfrac{9}{25}+\dfrac{1}{16}$. Multiplying through by $400$: $48x+25y=144+25=169$.
Correct Answer: 1