JEE 2023 Mathematics PYQ — Let the tangent to the parabola at the point be perpendicular to … | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the tangent to the parabola y2=12x at the point (3,α) be perpendicular to the line 2x+2y=3. Then, the square of distance of the point (6,−4) from the normal to the hyperbola α2x2−9y2=9α2 at its point (α−1,α+2) is equal to _____ .
Choose the correct answer:
A.
116
(Correct Answer)
B.
117
C.
118
D.
119
Correct Answer:
116
Explanation
Step 1: α ki value nikalna
Point (3,α) parabola y2=12x par hai, isliye:
α2=12(3)=36⟹α=±6
Ab, tangent ki slope (m1) nikalte hain:
2ydxdy=12⟹dxdy=y6
Point (3,α) par, m1=α6.
Di gayi line 2x+2y=3 ki slope (m2) hai −1.
Kyunki tangent perpendicular hai, toh m1×m2=−1:
(α6)(−1)=−1⟹α=6.
Step 2: Hyperbola aur Normal ki Equation
α=6 ko hyperbola ki equation mein rakhein:
62x2−9y2=9(62)⟹36x2−9y2=324
Dividing by 324: 9x2−36y2=1 (Yahan a2=9,b2=36).
Point (α−1,α+2) ho jayega (5,8).
Hyperbola a2x2−b2y2=1 ke point (x1,y1) par normal ki equation hoti hai:
x1a2x+y1b2y=a2+b2
Values rakhein:
59x+836y=9+36
59x+29y=45
Pure equation ko 9 se divide karein: 5x+2y=5
L.C.M lene par: 2x+5y−50=0 (Ye normal ki equation hai).
Step 3: Distance Calculation
Humein point (6,−4) se normal 2x+5y−50=0 ki distance (d) nikalni hai:
d=22+522(6)+5(−4)−50=2912−20−50=29−58
d=2958=229
Humein distance ka square nikalna hai:
d2=(229)2=4×29=116
Final Answer:
Square of the distance is 116.
Explanation
Step 1: α ki value nikalna
Point (3,α) parabola y2=12x par hai, isliye:
α2=12(3)=36⟹α=±6
Ab, tangent ki slope (m1) nikalte hain:
2ydxdy=12⟹dxdy=y6
Point (3,α) par, m1=α6.
Di gayi line 2x+2y=3 ki slope (m2) hai −1.
Kyunki tangent perpendicular hai, toh m1×m2=−1:
(α6)(−1)=−1⟹α=6.
Step 2: Hyperbola aur Normal ki Equation
α=6 ko hyperbola ki equation mein rakhein:
62x2−9y2=9(62)⟹36x2−9y2=324
Dividing by 324: 9x2−36y2=1 (Yahan a2=9,b2=36).
Point (α−1,α+2) ho jayega (5,8).
Hyperbola a2x2−b2y2=1 ke point (x1,y1) par normal ki equation hoti hai:
x1a2x+y1b2y=a2+b2
Values rakhein:
59x+836y=9+36
59x+29y=45
Pure equation ko 9 se divide karein: 5x+2y=5
L.C.M lene par: 2x+5y−50=0 (Ye normal ki equation hai).
Step 3: Distance Calculation
Humein point (6,−4) se normal 2x+5y−50=0 ki distance (d) nikalni hai:
