Question 67

If the shortest distance of the parabola $$y^2 = 4x$$ from the centre of the circle $$x^2 + y^2 - 4x - 16y + 64 = 0$$ is $$d$$, then $$d^2$$ is equal to :

The parabola is $$y^2 = 4x$$ with points $$(t^2, 2t)$$. The circle is $$(x-2)^2 + (y-8)^2 = 4$$ with centre $$C(2, 8)$$.

Squared distance from $$C$$ to a point on the parabola:

$$f(t) = (t^2 - 2)^2 + (2t - 8)^2 = t^4 - 4t^2 + 4 + 4t^2 - 32t + 64 = t^4 - 32t + 68$$

$$f'(t) = 4t^3 - 32 = 0 \implies t = 2$$

$$f(2) = 16 - 64 + 68 = 20$$

So $$d^2 = 20$$.

The correct answer is Option 3: $$20$$.

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