Maximum number of common normals of y2=4 a x and x2=4 b y may be equal to

Question

Maximum number of common normals of y2=4 a x and x2=4 b y may be equal to (A) 3 (B) 5 (C) 4 (D) none of these

Tardigrade Admin · Accepted Answer

Equations of normals to y2=4 a x and x2=4 b y are given by y=m x-2 a m-a m3 and y=m x+2 b+(b/m2) For common normals, 2 b+(b/m2)+2 a m+a m3=0 ⇒ a m5+2 a m3+2 b m2+b=0 So, a maximum of 5 normals are possible.

Q. Maximum number of common normals of $y^{2} = 4 a x$ and $x^{2} = 4 b y$ may be equal to

3

5

4

none of these

Equations of normals to $y^{2} = 4 a x$ and $x^{2} = 4 b y$ are given by
$y = m x - 2 am - a m^{3}$ and $y = m x + 2 b + \frac{b}{m ^{2}}$
For common normals, $2 b + \frac{b}{m ^{2}} + 2 am + a m^{3} = 0$
$\Rightarrow a m^{5} + 2 a m^{3} + 2 b m^{2} + b = 0$
So, a maximum of 5 normals are possible.

Q. Maximum number of common normals of y2=4ax and x2=4by may be equal to

3

5

4

none of these

Equations of normals to y2=4ax and x2=4by are given by y=mx−2am−am3 and y=mx+2b+m2b​ For common normals, 2b+m2b​+2am+am3=0 ⇒am5+2am3+2bm2+b=0 So, a maximum of 5 normals are possible.

Q. Maximum number of common normals of $y^{2} = 4 a x$ and $x^{2} = 4 b y$ may be equal to

Equations of normals to $y^{2} = 4 a x$ and $x^{2} = 4 b y$ are given by
$y = m x - 2 am - a m^{3}$ and $y = m x + 2 b + \frac{b}{m ^{2}}$
For common normals, $2 b + \frac{b}{m ^{2}} + 2 am + a m^{3} = 0$
$\Rightarrow a m^{5} + 2 a m^{3} + 2 b m^{2} + b = 0$
So, a maximum of 5 normals are possible.