JEE 2023 — Mathematics PYQ

Diya gaya curve ek ellipse hai: $9x^2+16y^2=144$

Dono taraf $144$ se divide karne par:

Yahan $a^2=16⟹a=4$ aur $b^2=9⟹b=3$.

Step 2: Tangent ki equation (Parametric form)

Ellipse ke kisi bhi point $(a\cos \theta ,b\sin \theta )$ par tangent ki equation hoti hai:

Value rakhne par: $\frac{x\cos \theta }{4}+\frac{y\sin \theta }{3}=1$

Step 3: Points $A$ aur $B$ ke coordinates

• Point $A$ (x-axis par): $y=0$ rakhein, toh $x=\frac{4}{\cos \theta }$. Isliye $A=\left(\frac{4}{\cos \theta },0\right)$.

• Point $B$ (y-axis par): $x=0$ rakhein, toh $y=\frac{3}{\sin \theta }$. Isliye $B=\left(0,\frac{3}{\sin \theta }\right)$.

Step 4: Line segment $AB$ ki length ($L$)

Distance formula se:

Step 5: Minimum length nikalna

Hume pata hai ki $a^2{\sec }^2\theta +b^2{\csc }^2\theta$ ki minimum value $(a+b{)}^2$ hoti hai.

Yahan $a^2=16⟹a=4$ aur $b^2=9⟹b=3$.

Final Answer:

Line segment $AB$ ki minimum length 7 hai.