Original Question: Find the equation of the tangent at point P of parabola $x^2=4y$, which is at the minimum distance to center of a circle $x^2+y^2+6x+8=0$.
What i did:
Found coordinates of center of circle i.e. $(-3,0)$
Equation of tangent at point $P(h,k)$, $hx-2y-2k=0$.
Now how to proceed further?
You really don't need the distance formula to solve this problem. Just a clear and intuitive understanding of geometry. Remember that if you have a line in space, and a point not on that line, the shortest distance between point and line make a right angle. Because of this property, the shortest distance between the circle's center and the parabola should make a right angle. That is to say, the normal line of the parabola must pass through the circle's center.
So first, we must find the equation of the normal to the parabola $y=x^2/4$. Simple enough, we can use point-slope form: $y-y_0 = m(x-x_0)$ where we'll plug in our point $(a, a^2/4)$. The slope of a normal line is the inverse reciprocal of the tangent. $y' = \frac{x}2$ --> normal slope = $-\frac2x$.
Equation of normal line: $y-\frac{a^2}4 = -\frac2a(x-a)$.
This normal line at point $a$ must pass through the center of the circle, which you've found (correctly) to be $(-3,0)$. So... Plug and solve!
$$0-\frac{a^2}4=-\frac2a(-3-a)$$
Simplify$$-\frac{a^2}4=\frac6a+2$$I multiplied by 4a$$-a^3=24+8a$$ $$a^3+8a+24=0$$ This has one solution, at $a=2$. At that point is the minimum distance between the parabola and the center of the circle. On the parabola, this would be the point $(2,1)$. Now we can find the tangent line at that point! Remember, slope of the tangent line is $\frac{x}2$.$y-1=\frac22(x-2)$ ------> $y=x-1$
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