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Conic Sections: JEE Main Mathematics Question with Solution Let a tangent to the curve
9 x 2 + 16 y 2 = 144 9x^2 + 16y^2 = 144 9 x 2 + 16 y 2 = 144 intersect the coordinate axes at the points
A \mathrm{A} A and
B \mathrm{B} B . Then, the minimum length of the line segment
A B \mathrm{AB} AB is
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The minimum length of the line segment A B \mathrm{AB} AB 7 7 7 Option analysis
Why each option works or fails
Step Working
01 given The ellipse is 9 x 2 + 16 y 2 = 144 9x^2 + 16y^2 = 144 9 x 2 + 16 y 2 = 144 x 2 16 + y 2 9 = 1 \frac{x^2}{16} + \frac{y^2}{9} = 1 16 x 2 + 9 y 2 = 1 a = 4 a = 4 a = 4 b = 3 b = 3 b = 3 A A A B B B
02 goal Find the minimum length of the segment A B AB A B
03 approach Write the tangent in parametric form at point ( 4 cos θ , 3 sin θ ) (4\cos\theta, 3\sin\theta) ( 4 cos θ , 3 sin θ ) A A A B B B A B 2 = x B 2 + y A 2 AB^2 = x_B^2 + y_A^2 A B 2 = x B 2 + y A 2 θ \theta θ
04 execute The equation of the tangent at P ( 4 cos θ , 3 sin θ ) P(4\cos\theta, 3\sin\theta) P ( 4 cos θ , 3 sin θ ) x cos θ 4 + y sin θ 3 = 1 \frac{x\cos\theta}{4} + \frac{y\sin\theta}{3} = 1 4 x c o s θ + 3 y s i n θ = 1 B ( 4 sec θ , 0 ) B(4\sec\theta, 0) B ( 4 sec θ , 0 ) A ( 0 , 3 csc θ ) A(0, 3\csc\theta) A ( 0 , 3 csc θ ) A B AB A B L = ( 4 sec θ ) 2 + ( 3 csc θ ) 2 = 16 sec 2 θ + 9 csc 2 θ L = \sqrt{(4\sec\theta)^2 + (3\csc\theta)^2} = \sqrt{16\sec^2\theta + 9\csc^2\theta} L = ( 4 sec θ ) 2 + ( 3 csc θ ) 2 = 16 sec 2 θ + 9 csc 2 θ
05 execute Convert to tan 2 θ \tan^2\theta tan 2 θ cot 2 θ \cot^2\theta cot 2 θ 16 sec 2 θ + 9 csc 2 θ = 16 ( 1 + tan 2 θ ) + 9 ( 1 + cot 2 θ ) = 25 + 16 tan 2 θ + 9 cot 2 θ 16\sec^2\theta + 9\csc^2\theta = 16(1 + \tan^2\theta) + 9(1 + \cot^2\theta) = 25 + 16\tan^2\theta + 9\cot^2\theta 16 sec 2 θ + 9 csc 2 θ = 16 ( 1 + tan 2 θ ) + 9 ( 1 + cot 2 θ ) = 25 + 16 tan 2 θ + 9 cot 2 θ 25 + 16 tan 2 θ + 9 cot 2 θ = 25 + ( 4 tan θ − 3 cot θ ) 2 + 2 ⋅ 4 ⋅ 3 = 25 + 24 + ( 4 tan θ − 3 cot θ ) 2 = 49 + ( 4 tan θ − 3 cot θ ) 2 25 + 16\tan^2\theta + 9\cot^2\theta = 25 + (4\tan\theta - 3\cot\theta)^2 + 2 \cdot 4 \cdot 3 = 25 + 24 + (4\tan\theta - 3\cot\theta)^2 = 49 + (4\tan\theta - 3\cot\theta)^2 25 + 16 tan 2 θ + 9 cot 2 θ = 25 + ( 4 tan θ − 3 cot θ ) 2 + 2 ⋅ 4 ⋅ 3 = 25 + 24 + ( 4 tan θ − 3 cot θ ) 2 = 49 + ( 4 tan θ − 3 cot θ ) 2 4 tan θ = 3 cot θ 4\tan\theta = 3\cot\theta 4 tan θ = 3 cot θ L min = 49 = 7 L_{\min} = \sqrt{49} = 7 L m i n = 49 = 7
✓ verify For any ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 a + b a + b a + b a = 4 a = 4 a = 4 b = 3 b = 3 b = 3 a + b = 4 + 3 = 7 a + b = 4 + 3 = 7 a + b = 4 + 3 = 7
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Question type Numerical
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Can we directly apply AM-GM on 16 sec 2 θ 16\sec^2\theta 16 sec 2 θ 9 csc 2 θ 9\csc^2\theta 9 csc 2 θ No, because their product depends on θ \theta θ sec 2 θ csc 2 θ = 4 sin 2 2 θ \sec^2\theta\csc^2\theta = \frac{4}{\sin^2 2\theta} sec 2 θ csc 2 θ = s i n 2 2 θ 4 tan 2 θ \tan^2\theta tan 2 θ cot 2 θ \cot^2\theta cot 2 θ
Answer The minimum length of the line segment A B \mathrm{AB} AB 7 7 7
Why each option works or fails Step-by-step solution given: The ellipse is 9 x 2 + 16 y 2 = 144 9x^2 + 16y^2 = 144 9 x 2 + 16 y 2 = 144 x 2 16 + y 2 9 = 1 \frac{x^2}{16} + \frac{y^2}{9} = 1 16 x 2 + 9 y 2 = 1 a = 4 a = 4 a = 4 b = 3 b = 3 b = 3 A A A B B B goal: Find the minimum length of the segment A B AB A B approach: Write the tangent in parametric form at point ( 4 cos θ , 3 sin θ ) (4\cos\theta, 3\sin\theta) ( 4 cos θ , 3 sin θ ) A A A B B B A B 2 = x B 2 + y A 2 AB^2 = x_B^2 + y_A^2 A B 2 = x B 2 + y A 2 θ \theta θ execute: The equation of the tangent at P ( 4 cos θ , 3 sin θ ) P(4\cos\theta, 3\sin\theta) P ( 4 cos θ , 3 sin θ ) x cos θ 4 + y sin θ 3 = 1 \frac{x\cos\theta}{4} + \frac{y\sin\theta}{3} = 1 4 x c o s θ + 3 y s i n θ = 1 B ( 4 sec θ , 0 ) B(4\sec\theta, 0) B ( 4 sec θ , 0 ) A ( 0 , 3 csc θ ) A(0, 3\csc\theta) A ( 0 , 3 csc θ ) A B AB A B L = ( 4 sec θ ) 2 + ( 3 csc θ ) 2 = 16 sec 2 θ + 9 csc 2 θ L = \sqrt{(4\sec\theta)^2 + (3\csc\theta)^2} = \sqrt{16\sec^2\theta + 9\csc^2\theta} L = ( 4 sec θ ) 2 + ( 3 csc θ ) 2 = 16 sec 2 θ + 9 csc 2 θ execute: Convert to tan 2 θ \tan^2\theta tan 2 θ cot 2 θ \cot^2\theta cot 2 θ 16 sec 2 θ + 9 csc 2 θ = 16 ( 1 + tan 2 θ ) + 9 ( 1 + cot 2 θ ) = 25 + 16 tan 2 θ + 9 cot 2 θ 16\sec^2\theta + 9\csc^2\theta = 16(1 + \tan^2\theta) + 9(1 + \cot^2\theta) = 25 + 16\tan^2\theta + 9\cot^2\theta 16 sec 2 θ + 9 csc 2 θ = 16 ( 1 + tan 2 θ ) + 9 ( 1 + cot 2 θ ) = 25 + 16 tan 2 θ + 9 cot 2 θ 25 + 16 tan 2 θ + 9 cot 2 θ = 25 + ( 4 tan θ − 3 cot θ ) 2 + 2 ⋅ 4 ⋅ 3 = 25 + 24 + ( 4 tan θ − 3 cot θ ) 2 = 49 + ( 4 tan θ − 3 cot θ ) 2 25 + 16\tan^2\theta + 9\cot^2\theta = 25 + (4\tan\theta - 3\cot\theta)^2 + 2 \cdot 4 \cdot 3 = 25 + 24 + (4\tan\theta - 3\cot\theta)^2 = 49 + (4\tan\theta - 3\cot\theta)^2 25 + 16 tan 2 θ + 9 cot 2 θ = 25 + ( 4 tan θ − 3 cot θ ) 2 + 2 ⋅ 4 ⋅ 3 = 25 + 24 + ( 4 tan θ − 3 cot θ ) 2 = 49 + ( 4 tan θ − 3 cot θ ) 2 4 tan θ = 3 cot θ 4\tan\theta = 3\cot\theta 4 tan θ = 3 cot θ L min = 49 = 7 L_{\min} = \sqrt{49} = 7 L m i n = 49 = 7 verify: For any ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 a + b a + b a + b a = 4 a = 4 a = 4 b = 3 b = 3 b = 3 a + b = 4 + 3 = 7 a + b = 4 + 3 = 7 a + b = 4 + 3 = 7 Shortcut: When to use it: Direct standard result: for an ellipse x 2 / a 2 + y 2 / b 2 = 1 x^2/a^2 + y^2/b^2 = 1 x 2 / a 2 + y 2 / b 2 = 1 a + b a + b a + b
given: Ellipse is x 2 4 2 + y 2 3 2 = 1 \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1 4 2 x 2 + 3 2 y 2 = 1 a = 4 a = 4 a = 4 b = 3 b = 3 b = 3
goal: Find the minimum intercept length A B AB A B
approach: Use the standard geometric theorem: the minimum length of a tangent segment between the coordinate axes for an ellipse x 2 / a 2 + y 2 / b 2 = 1 x^2/a^2 + y^2/b^2 = 1 x 2 / a 2 + y 2 / b 2 = 1 a + b a + b a + b
execute: L min = a + b = 4 + 3 = 7 L_{\min} = a + b = 4 + 3 = 7 L m i n = a + b = 4 + 3 = 7
verify: The condition holds for tan 2 θ = b / a = 3 / 4 \tan^2\theta = b/a = 3/4 tan 2 θ = b / a = 3/4 θ \theta θ