Test: Conic Sections: Hyerbola (22 July) - JEE MCQ

4x²−9y²−8x=32
⇒4(x²−2x)−9y2 = 32
⇒4(x²−2x+1)−9y² = 32 + 4 = 36
⇒(x−1)²]/9 − [y2]/4 = 1
⇒a²=9, b²=4
∴e=[1+b²/a²]^1/2 
= [(13)^1/2]/3

Given equation of line are
√3x−y−4√3k=0 …(i)
and √3kx+ky−4√3=0
From Eq. (i) 4√3–√k=3–√x−y
⇒ k=(√3x−y)/4√3
put in Eq. (ii), we get
√3x(√3x−y)/4√3)+((√3x−y)/4√3)y−4√3=0
⇒1/4(√3x²−xy)+1/4(xy−y²/√3)−4√3=0
⇒√3/4x²−y²/4√3-4√3=0
⇒3x²−y²−48=0
⇒3x²−y²=48,which is hyperbola.

Give eccentricity of the hyperbola is, e= 3/(5)^1/2
​⇒ (b²)/(a²) = 4/5..(1)
And latus rectum is =2(b²)/a=8
⇒ b²/a=4…(2)
By (2)/(1) a=5
∴ b²=20
Hence required hyperbola is, (x²)/25−(y²)/20=1
⇒ 4x²−5y² =100

Let the centre of hyperbola be (α,β)
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)²]/a² − [(y−β)²]/b² = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b² = a²(e² − 1)
⇒ b²=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)²]/16−[(y−5)²]/9=1

Polar is xy₁ + x₁y = 2c².. (1)
Let  normal chord at (h,k) be
hx - ky = h² - k²...... (2)
From 1 and
And hk = c² eliminate h, k and λ

hx + ky - xy = 0
let asymptotes be hx + ky - xy + c = 0 ; This represents a pair of straight lines if Δ =  0
i.e. c =-hk ∴ asymptotes are xh+yk-xy-hk =0 ⇒ (x - k) (y - h) = 0

Pt of intersection of x-y-1=0 and 3x-4y-6=0 is (-2,-3) other asymptote will be in the form of 4x + 3y + λ =0 and it should pass through (-2,-3). Thus λ = -17

 x²/2 - y²/3 = 1
= [y² + (3)2]/[x² - (2)²] = 1
=> [y² + 9]/[x² - 4] = 1
=> [y² + 9] = [x² - 4]
x² - y² = 13

Slope of 

 ------(1)
now solving the curves

-------(2)
from (1) & (2)

x² – 5xy + 6y² = 0 represent a pair of lines passing through origin
ax² + ay² = - c 

⇒ ax² + ay² + c = 0 represent a circle