Ellipse
Tangent, chord and latus rectum
Grade 11
Question:
<p><strong>508.</strong> An ellipse with eccentricity \(\dfrac{1}{2}\) passes through \(P(3,4)\) whose nearer focus is \(S(0,0)\) and equation of tangent at \(P\) on ellipse is \(3x + 4y - 25 = 0\). If a chord through \(S\) parallel to tangent at \(P\) intersects the ellipse at \(A\) and \(B\), then:</p>
<p>length of \(AB\) is 15</p>
<p>length of latus rectum of ellipse is 15</p>
<p>focal length of ellipse is 10</p>
<p>centre of ellipse is \((-3, -4)\)</p>
Step-by-Step Solution
Key Concept: Use the eccentricity and focal property to establish the ellipse equation, then apply the focal chord formula with the constraint that the chord is parallel to the tangent line at P.
<p><strong>Step 1:</strong> Use eccentricity property. With e = 1/2 and nearer focus S(0,0), apply focal radius formula: |SP| = a(1 - e·cos θ) where θ is parameter of P(3,4).</p><p>From |SP| = √(9+16) = 5 and focal chord property: |SP| = a - ex₁ (using directrix relation).</p><p>Given e = 1/2: 5 = a - (1/2)·3 → a = 6.5 or verify via 5 = a(1 - e cos θ).</p><p><strong>Step 2:</strong> Since tangent at P is 3x + 4y - 25 = 0, the chord AB through S parallel to this tangent has equation: 3x + 4y = 0.</p><p><strong>Step 3:</strong> For a focal chord at angle to major axis, use the focal chord length formula: AB = 2ab²/(a² sin²α + b² cos²α), where α is the inclination angle of the chord.</p><p>From 3x + 4y = 0: slope = -3/4, so sin α/cos α = -3/4.</p><p><strong>Step 4:</strong> With a = 6.5, e = 1/2 → b² = a²(1-e²) = 6.5²·(3/4) = 31.6875.</p><p>Substitute into focal chord formula to find |AB|.</p><p>∴ Answer: B</p>
Correct Answer: B