Conic Sections

For the parabola y = x², the tangent at (2,4) is given by (y+4)/2 = 2x, which simplifies to 4x - y - 4 = 0.
The equation of a circle touching the line 4x - y - 4 = 0 at the point (2,4) is
(x-2)² + (y-4)² + λ (4x-y-4) = 0.
It passes through (0,1).
∴ (0-2)² + (1-4)² + λ (4 (0) - 1 - 4) = 0
4 + 9 + λ (-5) = 0 ⇒ 13 = 5λ ⇒ λ = 13/5
∴ the circle is x² - 4x + 4 + y² - 8y + 16 + (13/5) (4x-y-4) = 0
x² + y² + (-4 + 52/5)x + (-8 - 13/5)y + (20 - 52/5) = 0
x² + y² + (32/5)x - (53/5)y + 48/5 = 0
∴ centre is (-g, -f) = (-16/5, 53/10).

Equation of normal to the ellipse x²/a² + y²/b² = 1 at (x? , y? ) is a²x/x? - b²y/y? = a² - b².
At the point (ae, b²/a):
a²x/ (ae) - b²y/ (b²/a) = a² - b²
It passes through (0, -b).
a² (0)/ (ae) - b² (-b)/ (b²/a) = a² - b²
ab = a² - b²
Since b² = a² (1-e²), a²-b² = a²e².
ab = a²e²
a²b² = a? e?
a² (a² (1-e²) = a? e?
1 - e² = e?
e? + e² - 1 = 0

Given : a/e = 4 and 1/4 = 1 - b²/a²
Solving : a = 2, b = √3
Parametric co - ordinates are
(2cosθ, √3sinθ) = (1, β)
∴ θ = 60°
∴ Equation of normal is axsecθ − bycosecθ = a² − b²
⇒ 4x - 2y = 1

By family of circle, passing through intersection of given circle will be member of S + λS? = 0 family (λ ≠ 1)
(x² + y² – 6x) + λ (x² + y² – 4y) = 0
(λ + 1)x² + (λ + 1)y² – 6x – 4λy = 0
x² + y² - 6/ (λ+1) x - 4λ/ (λ+1) y = 0
Centre (3/ (λ+1), 2λ/ (λ+1)
Centre lies on 2x – 3y + 12 = 0
2 (3/ (λ+1) - 3 (2λ/ (λ+1) + 12 = 0
6λ + 18 = 0
λ = -3
Circle x² + y² – 3x – 6y = 0

Since (3,3) lies on x²/a² - y²/b² = 1
9/a² - 9/b² = 1
Now, normal at (3,3) is y-3 = -a²/b² (x-3),

which passes through (9,0) ⇒ b² = 2a²

So, e² = 1 + b²/a² = 3
Also, a² = 9/2
(From (i) and (ii)
Thus, (a², e²) = (9/2, 3)

 x²/a² + y²/b² = 1 (a > b); 2b²/a = 10 ⇒ b² = 5a
Now, φ (t) = -5/12 + t - t² = 8/12 - (t - 1/2)²
φ (t)max = 8/12 = 2/3 = e ⇒ e² = 1 - b²/a² = 4/9
⇒ a² = 81 (From (i) and (ii)
So, a² + b² = 81 + 45 = 126