Ellipse
Tangent and Normal
Grade 11
Question:
<p>If the tangent drawn at point \((t^2, 2t)\) on the parabola \(y^2 = 4x\) is same as the normal drawn at point \(\left(\sqrt{5}\cos\theta, 2\sin\theta\right)\) on the ellipse \(4x^2 + 5y^2 = 20\). Then</p>
<p>\(\theta = \cos^{-1}\left(-\dfrac{1}{\sqrt{5}}\right)\)</p>
<p>\(\theta = \cos^{-1}\left(\dfrac{1}{\sqrt{5}}\right)\)</p>
<p>\(t = -\dfrac{2}{\sqrt{5}}\)</p>
<p>\(t = -\dfrac{1}{\sqrt{5}}\)</p>
Step-by-Step Solution
Key Concept: The tangent at a parabola point and normal at an ellipse point are the same line—equate their slopes and use point conditions to find the relationship between parameters t and θ.
<p><strong>Step 1: Find tangent at (t², 2t) on parabola y² = 4x</strong></p><p>For parabola y² = 4x, the tangent at (t², 2t) is: <strong>ty = x + t²</strong></p><p>Slope of tangent: m = 1/t</p><p><strong>Step 2: Find normal at (√5cosθ, 2sinθ) on ellipse 4x² + 5y² = 20</strong></p><p>Standard form: x²/5 + y²/4 = 1, so a² = 5, b² = 4</p><p>Normal at (√5cosθ, 2sinθ) has slope: <strong>m = (a²sinθ)/(b²cosθ) = (5sinθ)/(4cosθ)</strong></p><p><strong>Step 3: Equate slopes</strong></p><p>1/t = 5sinθ/(4cosθ) ... (i)</p><p><strong>Step 4: Use point condition - tangent passes through (√5cosθ, 2sinθ)</strong></p><p>ty = x + t²</p><p>t(2sinθ) = √5cosθ + t²</p><p>2tsinθ - √5cosθ = t² ... (ii)</p><p><strong>Step 5: Solve equations (i) and (ii)</strong></p><p>From (i): 4cosθ = 5tsinθ</p><p>From (ii): 2tsinθ = √5cosθ + t²</p><p>Substituting 4cosθ = 5tsinθ → cosθ = 5tsinθ/4 into (ii):</p><p>2tsinθ = √5(5tsinθ/4) + t²</p><p>2tsinθ = 5√5tsinθ/4 + t²</p><p>∴ Answer: A</p>
Correct Answer: A