JEE Main 2021 · 16 March · morning · Single correct · +4 / −1
The locus of the midpoints of the chord of the circle, x² + y² = 25 which is tangent to the hyperbola, x²/9 - y²/16 = 1 is :
- (a) (x² + y²)² - 9x² + 16y² = 0
- (b) (x² + y²)² - 9x² + 144y² = 0
- (c) (x² + y²)² - 16x² + 9y² = 0
- (d) (x² + y²)² - 9x² - 16y² = 0
Answer and solution
Answer (a) (x² + y²)² - 9x² + 16y² = 0
Solution
Maths sir · 3 steps
Write the chord using its midpoint with T = S₁, then force that line to touch the hyperbola.
- Chord from midpoint. Let the midpoint be (h, k). For the circle x² + y² = 25, the chord with this midpoint is T = S₁, that is hx + ky - 25 = h² + k² - 25, so the chord is hx + ky = h² + k².
- Tangency condition. The line y = mx + c touches x²/a² - y²/b² = 1 exactly when c² = a²m² - b². Here a² = 9, b² = 16, and the chord has m = -h/k, c = (h² + k²)/k.
(h² + k²)²/k² = 9h²/k² - 16 ⇒ (h² + k²)² = 9h² - 16k²
- So. Replace (h, k) by (x, y): the locus is (x² + y²)² - 9x² + 16y² = 0, which is option (a).
The idea to keep. Locus of midpoints of chords: write the chord as T = S₁ in terms of (h, k), then apply the condition the chord must satisfy. For a hyperbola tangent, use c² = a²m² - b² with a minus sign.
Answer from the NTA final answer key. Solution by PrepWiser.