Test: Hyperbola- 1 - 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

Equation of line perpendicular to x−y=0 is given by
y=−x+c
Also this line is tangent to the hyperbola x²−2y²=18
So we have m=−1, a²=18, b²=9
Thus Using condition of tangency c² = a²m²−b²
= 18−9=9
⇒ c = ±3
Hence required equation of tangent is x+y = ±3

Given hyperbola is 25x²−16y²=400
If (6, 2) is the midpoint of the chord, then equation of chord is T = S1
​⇒25(6x)−16(2y)=25(36)−16(4)
⇒75x−16y=450−32
⇒75x−16y=418

xy - 3x - 2y + λ = 0.
Then abc + 2fgh − af² − bg² − ch² = 0
⇒ 3/2 − λ/4 = 0
⇒ λ = 6
∴ Equation of asymptotes is xy-3x-2y+6=0
⇒ (x-2)(y-3)=0
⇒x - 2 = 0 and y - 3 = 0

e² = (a² / b²) + 1
3 = (a² / b²) + 1
a²/b² = 2
a² = 2b²
Product of perpendicular distance of any point on hyperbola = (a²b²)/a² + b²
= [2(b²).b²]/(2b² + b²)
= [2b²]/3 = 6
= 2b² = 18
=> b² = 9
=> b = 3
Length (2b) = 2(3) 
= 6

Differentiating xy=16, (xdy)/x+y=0
⇒ dy/dx=−y/x=−1/16= Slope of tangent
⇒  Its (tangent's) equation : y=−1=−1/16(x−16)
⇒ 16y−16=−x+16 ⇒ 16y=−x−32
⇒ 16y+x+32=0
It will cut x−axis at A(−32, 0)
& y−axis at B(0, −2)
⇒  Area of △OAB= 1/2×2×32 
⇒ 32

Let middle point of chord is P(h,k)
Thus equation of chord with mid point P is,  kx+hy=2hk
Given slope of this chord is m
⇒ −k/h = m
⇒k+mh=0
Thus locus of P(h,k) is, y+mx=0