If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2},$ then the value of $r^{2}$ is

Open in App

Solution

$y^{2} = 4 x$
$Here a = 1,$
End points of letus rectum
$= (a, \pm 2 a) = (1, \pm 2)$
Comparing it with $(t^{2}, 2 t)$ , we get
$t = \pm 1$
Hence slope of normal is
$m = - t = \mp 1$
The equation of normal is
$y + t x = 2 a t + a t^{3}$
So, normal are
$x + y = 3 and x - y = 3$

These are also tangents to the circle
$(x - 3)^{2} + (y + 2)^{2} = r^{2}$
Perpendicular distance from the centre is equal to radius
$\Rightarrow ∣ ∣ ∣ \frac{3 \mp 2 - 3}{\sqrt{1 + 1}} ∣ ∣ ∣ = r ∴ r^{2} = 2$

Suggest Corrections

Similar questions

If the normals of the parabola y²=4x drawn at the end points of its latus rectum are tangents to the circle (x-3)²+(y+2)²=r², then the value of r² is

If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

Q. Two tangents are drawn to end points of the latus rectum of the parabola

y^{2} = 4 x

. The equation of the parabola which touches both the tangents as well as the latus rectum is

Q. Two tangents are drawn to end points of the latus rectum of the parabola

y^{2} = 4 x

. The equation of the parabola which touches both the tangents as well as the latus rectum is

Q. The point of intersection of the normals to the parabola

y^{2} = 4 x

at the ends of its latus rectum is :

Join BYJU'S Learning Program

Explore more

Normal

Standard XII Mathematics

Join BYJU'S Learning Program

If the normals of the parabola y2=4x drawn at the end points of its latus rectum are tangents to the circle (x–3)2+(y+2)2=r2, then the value of r2 is

If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2},$ then the value of $r^{2}$ is