NIMCET 2014 Mathematics PYQ — The locus of the intersection of the two lines and , for differen… | Mathem Solvex | Mathem Solvex
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NIMCET 2014 — Mathematics PYQ
NIMCET | Mathematics | 2014
The locus of the intersection of the two lines 3x−y=4k3 and k(3x+y)=43, for different values of k, is a hyperbola. The eccentricity of the hyperbola is:
Choose the correct answer:
A.
1.5
B.
3
C.
2
(Correct Answer)
D.
23
Correct Answer:
2
Explanation
Solution
Concept:
Two curves f(x,y)=0 and g(x,y)=0 cut/touch at a point (a,b) if f(a,b)=g(a,b)=0.
The eccentricity (e) of the hyperbola a2x2−b2y2=1 is given by:
e=ac=1+a2b2
Calculation:
Let the equations of the lines be:
f(x,y)=3x−y−4k3=0
g(x,y)=k3x+ky−43=0
If f(x,y)=0 and g(x,y)=0 intersect at a point (a,b), then we must have:
f(a,b)=g(a,b)=0
⇒3a−b−4k3=k3a+kb−43=0
Multiplying the first expression by −k gives us:
⇒−k3a+kb+4k23=k3a+kb−43=0
⇒2k3a=4k23+43
⇒ka=2k2+2
⇒a=2(kk2+1)
Substituting this in f(a,b)=0:
⇒b=3(a−4k)
⇒b=3(k2k2+2−4k)
⇒b=23(k1−k2)
It can be observed that:
(a3)2−b2=[23(kk2+1)]2−[23(k1−k2)]2
⇒3a2−b2=12(k2k4+2k2+1−k4+2k2−1)
⇒3a2−b2=48
The locus of the point of intersection (a,b) is:
3x2−y2=48
⇒16x2−48y2=1
Comparing this with the general equation of a hyperbola a2x2−b2y2=1 and using the eccentricity formula:
e=1+1648=1+3=2
Correct Option: 3. (2)
Explanation
Solution
Concept:
Two curves f(x,y)=0 and g(x,y)=0 cut/touch at a point (a,b) if f(a,b)=g(a,b)=0.
The eccentricity (e) of the hyperbola a2x2−b2y2=1 is given by:
e=ac=1+a2b2
Calculation:
Let the equations of the lines be:
f(x,y)=3x−y−4k3=0
g(x,y)=k3x+ky−43=0
If f(x,y)=0 and g(x,y)=0 intersect at a point (a,b), then we must have:
f(a,b)=g(a,b)=0
⇒3a−b−4k3=k3a+kb−43=0
Multiplying the first expression by −k gives us:
⇒−k3a+kb+4k23=k3a+kb−43=0
⇒2k3a=4k23+43
⇒ka=2k2+2
⇒a=2(kk2+1)
Substituting this in f(a,b)=0:
⇒b=3(a−4k)
⇒b=3(k2k2+2−4k)
⇒b=23(k1−k2)
It can be observed that:
(a3)2−b2=[23(kk2+1)]2−[23(k1−k2)]2
⇒3a2−b2=12(k2k4+2k2+1−k4+2k2−1)
⇒3a2−b2=48
The locus of the point of intersection (a,b) is:
3x2−y2=48
⇒16x2−48y2=1
Comparing this with the general equation of a hyperbola a2x2−b2y2=1 and using the eccentricity formula:
e=1+1648=1+3=2
Correct Option: 3. (2)
NIMCET
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