The equations of the normals at the ends of the latusrectum of the parabola y = 4ax are given by

Question

The equations of the normals at the ends of the latusrectum of the parabola y = 4ax are given by (A)  x2 - y2 - 6ax + 9a2 = 0  (B)  x2 - y2 - 6ax - 6ay + 9a2 = 0  (C)  x2 - y2 - 6ay + 9a2 = 0  (D) None of the above

Tardigrade Admin · Accepted Answer

The coordinates of the ends of the latusrectum of the parabola

y^{2} = 4 a x

are

(a, 2 a)

respectively. The equation of the normal at

(a, 2 a)

to

y^{2} = 4 a x

is

y - 2 a = \frac{- 2 a}{2 a} (x - a)

or

x + y - 3 a = 0

Similarly, the equation of the normal

(a, - 2 a)

is

x - y - 3 a = 0

The combined equation is

x^{2} - y^{2} - 6 a x + 9 a^{2} = 0

Q. The equations of the normals at the ends of the latusrectum of the parabola $y = 4 a x$ are given by

$x^{2} - y^{2} - 6 a x + 9 a^{2} = 0$

$x^{2} - y^{2} - 6 a x - 6 a y + 9 a^{2} = 0$

$x^{2} - y^{2} - 6 a y + 9 a^{2} = 0$

None of the above

Q. The equations of the normals at the ends of the latusrectum of the parabola y=4ax are given by

x2−y2−6ax+9a2=0

x2−y2−6ax−6ay+9a2=0

x2−y2−6ay+9a2=0

None of the above

The coordinates of the ends of the latusrectum of the parabola y2=4ax are (a,2a) respectively. The equation of the normal at (a,2a) to y2=4ax is y−2a=2a−2a​(x−a) or x+y−3a=0 Similarly, the equation of the normal (a,−2a) is x−y−3a=0 The combined equation is x2−y2−6ax+9a2=0

Q. The equations of the normals at the ends of the latusrectum of the parabola $y = 4 a x$ are given by

$x^{2} - y^{2} - 6 a x + 9 a^{2} = 0$

$x^{2} - y^{2} - 6 a x - 6 a y + 9 a^{2} = 0$

$x^{2} - y^{2} - 6 a y + 9 a^{2} = 0$