Question:
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is \(\frac{1}{2}\). If P(1, \(\beta\)), \(\beta \gt 0\) is a point on this ellipse, then the equation of the normal to it at P is :
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is \(\frac{1}{2}\). If P(1, \(\beta\)), \(\beta \gt 0\) is a point on this ellipse, then the equation of the normal to it at P is :
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Directrix \(x = a/e\) gives \(a\); then \(b^2 = a^2(1-e^2)\). Find \(\beta\) from the ellipse equation and use the normal formula at \((x_1, y_1)\).
Updated On: Nov 5, 2026
- \(4x - 3y = 2\)
- \(8x - 2y = 5\)
- \(7x - 4y = 1\)
- \(4x - 2y = 1\)
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The Correct Option is D
Solution and Explanation
Centre at origin, directrix \(x = \frac{a}{e} = 4\) and \(e = \frac{1}{2}\), so \(a = 2\).
\(b^2 = a^2(1-e^2) = 4\left(1-\frac{1}{4}\right) = 3\). Ellipse: \(\frac{x^2}{4} + \frac{y^2}{3} = 1\).
P(1, \(\beta\)) lies on it: \(\frac{1}{4} + \frac{\beta^2}{3} = 1 \Rightarrow \beta^2 = \frac{9}{4}\), and since \(\beta \gt 0\), \(\beta = \frac{3}{2}\).
Normal to \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at \((x_1, y_1)\): \(\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2\).
\(\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3 \Rightarrow 4x - 2y = 1\).
Answer: (4) 4x - 2y = 1
\(b^2 = a^2(1-e^2) = 4\left(1-\frac{1}{4}\right) = 3\). Ellipse: \(\frac{x^2}{4} + \frac{y^2}{3} = 1\).
P(1, \(\beta\)) lies on it: \(\frac{1}{4} + \frac{\beta^2}{3} = 1 \Rightarrow \beta^2 = \frac{9}{4}\), and since \(\beta \gt 0\), \(\beta = \frac{3}{2}\).
Normal to \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at \((x_1, y_1)\): \(\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2\).
\(\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3 \Rightarrow 4x - 2y = 1\).
Answer: (4) 4x - 2y = 1
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