Let P be a point on the parabola, x2 = 4y. If the distance of P from the centre of the circle, x2 + y2 + 6 x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :

Question

Let P be a point on the parabola, x2 = 4y. If the distance of P from the centre of the circle, x2 + y2 + 6 x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is : (A) x + 4y - 2 = 0 (B) x - y + 3 = 0 (C) x + y + 1 = 0 (D) x + 2y = 0

Tardigrade Admin · Accepted Answer

Let P(2 t, t2), equation normal at P to x2=4 y be y-t2=(-1/t)(x-2 t) It passes through (-3,0). 0-t2 =(-1/t)(-3-2 t) t3+2 t+3 =0 (t+1)(t2-t+3) =0 t =-1 So, point P is (-2,1). Therefore, equation of tangent to x2=4 y at (-2,1) is x(-2) =2(y+1) x+y+1 =0

Q. Let $P$ be a point on the parabola, $x^{2} = 4 y$ . If the distance of $P$ from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at $P$ , is :

$x + 4 y - 2 = 0$

$x - y + 3 = 0$

$x + y + 1 = 0$

$x + 2 y = 0$

Q. Let P be a point on the parabola, x2=4y. If the distance of P from the centre of the circle, x2+y2+6x+8=0 is minimum, then the equation of the tangent to the parabola at P, is :

x+4y−2=0

x−y+3=0

x+y+1=0

x+2y=0

Let P(2t,t2), equation normal at P to x2=4y be y−t2=t−1​(x−2t) It passes through (−3,0). 0−t2=t−1​(−3−2t) t3+2t+3=0 (t+1)(t2−t+3)=0 t=−1 So, point P is (−2,1). Therefore, equation of tangent to x2=4y at (−2,1) is x(−2)=2(y+1) x+y+1=0

Q. Let $P$ be a point on the parabola, $x^{2} = 4 y$ . If the distance of $P$ from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at $P$ , is :

$x + 4 y - 2 = 0$

$x - y + 3 = 0$

$x + y + 1 = 0$

$x + 2 y = 0$