Ellipse
Tangent and Normal to Ellipse
Grade 11
Question:
<p>If the line <span>\(x\cos\alpha + y\sin\alpha = p\)</span> be normal to the ellipse <span>\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)</span>, then</p>
<p>(a) <span>\(p^2(a^2\cos^2\alpha + b^2\sin^2\alpha) = a^2 - b^2\)</span></p>
<p>(b) <span>\(p^2(a^2\cos^2\alpha + b^2\sin^2\alpha) = (a^2 - b^2)^2\)</span></p>
<p>(c) <span>\(p^2(a^2\sec^2\alpha + b^2\cosec^2\alpha) = a^2 - b^2\)</span></p>
<p>(d) <span>\(p^2(a^2\sec^2\alpha + b^2\cosec^2\alpha) = (a^2 - b^2)^2\)</span></p>
Step-by-Step Solution
Key Concept: A line is normal to an ellipse if it is perpendicular to the tangent at the point of contact. Use the condition that the normal line must pass through a specific point on the ellipse, then eliminate the point of contact to derive the relationship between p, a, b, and α.
<p><strong>Step 1: Equation of normal to ellipse at parameter point</strong></p><p>For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the normal at point $(a\cos t, b\sin t)$ is:</p><p>$$\frac{ax}{\cos t} - \frac{by}{\sin t} = a^2 - b^2$$</p><p><strong>Step 2: Compare with given normal equation</strong></p><p>The given normal line is: $x\cos\alpha + y\sin\alpha = p$</p><p>Dividing the standard normal by $(a^2-b^2)$ and comparing:</p><p>$$\frac{ax}{\cos t(a^2-b^2)} - \frac{by}{\sin t(a^2-b^2)} = 1$$</p><p>For this to match $x\cos\alpha + y\sin\alpha = p$, we need:</p><p>$$\frac{a\cos\alpha}{\cos t(a^2-b^2)} = \frac{1}{p} \text{ and } \frac{-b\sin\alpha}{\sin t(a^2-b^2)} = \frac{1}{p}$$</p><p><strong>Step 3: Extract conditions</strong></p><p>From the above equations:</p><p>$$a\cos\alpha = \frac{\cos t(a^2-b^2)}{p}$$</p><p>$$-b\sin\alpha = \frac{\sin t(a^2-b^2)}{p}$$</p><p><strong>Step 4: Use the constraint $\cos^2 t + \sin^2 t = 1$</strong></p><p>Squaring both equations and adding:</p><p>$$a^2\cos^2\alpha + b^2\sin^2\alpha = \frac{(\cos^2 t + \sin^2 t)(a^2-b^2)^2}{p^2}$$</p><p>$$a^2\cos^2\alpha + b^2\sin^2\alpha = \frac{(a^2-b^2)^2}{p^2}$$</p><p><strong>Step 5: Rearrange to get the final form</strong></p><p>$$p^2(a^2\cos^2\alpha + b^2\sin^2\alpha) = (a^2-b^2)^2$$</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b