The tangent to the parabola y=x2-2x+8 at P(2 , 8) touches the circle x2+y2+18x+14y+λ =0 at Q. The coordinates of point Q are

Question

The tangent to the parabola y=x2-2x+8 at P(2 , 8) touches the circle x2+y2+18x+14y+λ =0 at Q. The coordinates of point Q are (A) (- 7 , - 12) (B) (- 9 , - 13) (C) (- 11 , - 16) (D) (- (31/5) , - (42/5))

Tardigrade Admin · Accepted Answer

The equation of the tangent at P (2 , 8) to the parabola is (y + 8/2)=2x-(x + 2)+8 ⇒ 2x-y+4=0 ...(.i.) So, the slope of the normal to the circle is -(1/2) The equation of normal to the circle at Q will be y+7=-(1/2)(x + 9) x+2y+23=0 ...(.ii.) The point Q is the intersection of the tangent at P and the normal at Q On solving the equation (.i.) (.ii.) , we get, x+2(2 x + 4)+23=0 ⇒ 5x=-31⇒ x=-(31/5) and y=2(- (31/5))+4=-(42/5) Hence, the coordinates of the point Q are (- (31/5) , - (42/5))

Q. The tangent to the parabola $y = x^{2} - 2 x + 8$ at $P (2, 8)$ touches the circle $x^{2} + y^{2} + 18 x + 14 y + λ = 0$ at $Q .$ The coordinates of point $Q$ are

$(- 7, - 12)$

$(- 9, - 13)$

$(- 11, - 16)$

$(- \frac{31}{5}, - \frac{42}{5})$

Q. The tangent to the parabola y=x2−2x+8 at P(2,8) touches the circle x2+y2+18x+14y+λ=0 at Q. The coordinates of point Q are

(−7,−12)

(−9,−13)

(−11,−16)

(−531​,−542​)

Q. The tangent to the parabola $y = x^{2} - 2 x + 8$ at $P (2, 8)$ touches the circle $x^{2} + y^{2} + 18 x + 14 y + λ = 0$ at $Q .$ The coordinates of point $Q$ are

$(- 7, - 12)$

$(- 9, - 13)$

$(- 11, - 16)$

$(- \frac{31}{5}, - \frac{42}{5})$