The tangent to the parabola y = x^2 + 6 at th | vedclass

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Question

(A) The equation of the tangent to the parabola $y = x^2 + 6$ at point $(1, 7)$ is given by $\frac{1}{2}(y + 7) = x(1) + 6$,which simplifies to $y = 2

Vedclass · Accepted Answer

$(-6, -7)$

Vedclass · Answer

(A) The equation of the tangent to the parabola $y = x^2 + 6$ at point $(1, 7)$ is given by $\frac{1}{2}(y + 7) = x(1) + 6$,which simplifies to $y = 2x + 5$ ... $(i)$This line is also a tangent to the circle $x^2 + y^2 + 16x + 12y + c = 0$ ... (ii)Substituting $y = 2x + 5$ into the circle equation:$x^2 + (2x + 5)^2 + 16x + 12(2x + 5) + c = 0$$x^2 + 4x^2 + 20x + 25 + 16x + 24x + 60 + c = 0$$5x^2 + 60x + (85 + c) = 0$ ... (iii)Since the line is tangent to the circle,the discriminant of this quadratic equation must be zero:$D = (60)^2 - 4(5)(85 + c) = 0$$3600 - 20(85 + c) = 0$$180 - (85 + c) = 0 \Rightarrow 85 + c = 180$Substituting $85 + c = 180$ back into equation (iii):$5x^2 + 60x + 180 = 0$$x^2 + 12x + 36 = 0$$(x + 6)^2 = 0 \Rightarrow x = -6$Using $y = 2x + 5$,we get $y = 2(-6) + 5 = -7$.Thus,the point of contact $Q$ is $(-6, -7)$.

The tangent to the parabola $y = x^2 + 6$ at the point $P(1, 7)$ touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ at point $Q$. What are the coordinates of $Q$?

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