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 + y + 1 = 0

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x + 4 y - 2 = 0

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x + 2 y = 0

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x - y + 3 = 0

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Solution

The correct option is A $x + y + 1 = 0$
$Let P (2 t, t^{2}) Centre of the circle : (- 3, 0) y = D^{2} = (2 t + 3)^{2} + (t^{2} - 0)^{2} For minimum distance, \frac{d y}{d t} = 0 \Rightarrow 2 (2 t + 3) 2 + 4 t^{3} = 0 \Rightarrow t^{3} + 2 t + 3 = 0 \Rightarrow (t + 1) (t^{2} - t + 3) = 0 \Rightarrow t = - 1 ∴ P is (- 2, 1) Equation of tangent to the parabola is, x x_{1} = 2 a (y + y_{1}) \Rightarrow x (- 2) = 2 \times 1 (y + 1) \Rightarrow x + y + 1 = 0$

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Similar questions

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

Q. Let P be the point on the parabola,

y^{2} = 8 x

, which is at a minimum distance from the centre C of the circle,

x^{2} + (y + 6)^{2} = 1

. Then, the equation of the circle, passing through C and having its centre at P is

Let P be the point on the parabola, $y^{2} = 8 x$ which is at a minimum distance from the centre C of the circle, $x^{2} + (y + 6)^{2} = 1.$ Then the equation of the circle, passing through C and having its centre at P is:

Q. The equation of common tangent to the parabolas

y^{2} = 4 x

and

x^{2} = 4 y

Q. Consider the parabola whose focus at

(0, 0)

and tangent at vertex is

x - y + 1 = 0

.
The equation of the parabola is

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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 :

The correct option is A x+y+1=0Let P(2t,t2)Centre of the circle : (−3,0)y=D2=(2t+3)2+(t2−0)2For minimum distance, dydt=0⇒2(2t+3)2+4t3=0⇒t3+2t+3=0⇒(t+1)(t2−t+3)=0⇒t=−1∴P is (−2,1)Equation of tangent to the parabola is, xx1=2a(y+y1)⇒x(−2)=2×1(y+1)⇒x+y+1=0

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 :