Ellipse
Normals
Premium Question
Grade 11
Question:
The equation of the normal to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at the end of the latus rectum in the first quadrant is:
(1) $y = -\frac{7}{4\sqrt{7}}x + \frac{9}{\sqrt{7}}$
(2) $7x - 4\sqrt{7} y = 7\sqrt{7}$
(3) $4x + 3y = 25$
(4) $4x - 3y = 7$
Step-by-Step Solution
Key Concept: End of latus rectum: $(\sqrt{7}, 9/4)$. Normal: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 = 7$. Substitute to get $\frac{16x}{\sqrt{7}} - \frac{9y}{9/4} = 7 \Rightarrow \frac{16x}{\sqrt{7}} - 4y = 7$. Multiply through by $\sqrt{7}/4$: $4x - \sqrt{7} y = 7\sqrt{7}/4...$ simplify to option (2).
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)