Explanation
1. Quadratic Curve ki Equation nikalna
Maan lijiye quadratic curve f(x)=ax2+bx+c hai. Hame teen conditions di gayi hain:
-
Condition 1: Curve (−1,0) se guzarta hai.
a(−1)2+b(−1)+c=0⟹a−b+c=0—(i)
-
Condition 2: Curve (1,1) se guzarta hai.
a(1)2+b(1)+c=1⟹a+b+c=1—(ii)
-
Condition 3: (1,1) par tangent ki slope y=x ke barabar hai (yani slope =1).
f′(x)=2ax+b. Isliye f′(1)=1:
Equations solve karne par:
(ii) mein se (i) ghatane par: 2b=1⟹b=1/2.
b ki value (iii) mein rakhne par: 2a+1/2=1⟹2a=1/2⟹a=1/4.
a aur b ki value (ii) mein rakhne par: 1/4+1/2+c=1⟹c=1/4.
Toh hamara curve hai: y=41x2+21x+41 ya 4y=(x+1)2.
2. Point (α,α+1) dhundna
Yeh point curve par sthit hai, isliye:
Dono taraf (α+1) se divide karne par (kyunki point first quadrant mein hai, toh α=−1):
Toh point hai (3,4).
3. Normal ki Equation aur x-intercept
Point (3,4) par curve ki slope (mt):
mt=f′(3)=2(1/4)(3)+1/2=3/2+1/2=2
Normal ki slope (mn) hogi:
Normal ki equation (3,4) par:
x-intercept ke liye y=0 rakhein:
