Q.Find the shortest distance of the point $(0, c)$ from the parabola $y = x^2$, where $\dfrac{1}{2} \le c \le 5$.

Concept and Intuition

For a convex curve like a parabola and an external line that doesn't intersect it, the shortest distance from the curve to the line occurs at the point where the curve's tangent line is parallel to the given line. This is because at that point, moving along the curve in either direction initially moves you no closer or farther (to first order) — it's the calculus analogue of "drop a perpendicular," generalized to curves.

Step-by-Step Solution

1. The line is $y=3x-2$, slope $=3$.
2. For the parabola $y=x^2+7x+2$, the slope of the tangent at any point is $\dfrac{dy}{dx}=2x+7$.
3. Set the tangent slope equal to the line's slope: $2x+7=3 \Rightarrow 2x=-4 \Rightarrow x=-2$.
4. Find $y$ at $x=-2$: $y=(-2)^2+7(-2)+2=4-14+2=-8$.
5. So the closest point is $(-2,-8)$. …

Concept and Intuition

For a point $(x,y,z)$, its perpendicular distance from the X-axis is $\sqrt{y^2+z^2}$, from the Y-axis is $\sqrt{x^2+z^2}$, and from the Z-axis is $\sqrt{x^2+y^2}$. Here $x=3,y=4$ are fixed, so the Z-axis distance is a constant, while the other two both involve $\alpha^2$ (via the free $z=\alpha$ coordinate) and are each individually minimized when $\alpha^2$ is as small as possible, i.e. $\alpha=0$.

Step-by-Step Solution

1. Point: $(3,4,\alpha)$.
2. Distance from X-axis $=\sqrt{y^2+z^2}=\sqrt{16+\alpha^2}$.
3. Distance from Y-axis $=\sqrt{x^2+z^2}=\sqrt{9+\alpha^2}$.
4. Distance from Z-axis $=\sqrt{x^2+y^2}=\sqrt{9+16}=5$ (constant, no $\alpha$-dependence).
5. Sum $=\sqrt{16+\alpha^2}+\sqrt{9+\alpha^2}+5$. …

Concept and Intuition

A line crossing both axes, together with the origin, forms a right triangle (the two axes are perpendicular, meeting at the origin). For any right triangle, the circumcircle's diameter is exactly the hypotenuse (Thales' theorem) — so the circumcenter is the hypotenuse's midpoint and the circumradius is half its length. Once we have circle $S$, the closest point $Q$ on it to an external point $P$ lies along the straight line from $P$ through the center, at distance (center-to-$P$ distance) minus radius.

Step-by-Step Solution

1. Intercepts of $x-2y-4=0$: setting $y=0$ gives $x=4$; setting $x=0$ gives $y=-2$. So the triangle has vertices $(0,0),(4,0),(0,-2)$.
2. Right angle is at the origin (the two legs lie along the axes), so the hypotenuse (from $(4,0)$ to $(0,-2)$) is the circle's diameter.
3. Circumcenter $=$ midpoint of $(4,0)$ and $(0,-2)$ $=(2,-1)$.
4. Hypotenuse length $=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt5$, so radius $r=\sqrt5$. …

Concept and Intuition

Distance is minimised where its square is minimised. Writing $D^2$ as a function of $t$ and finding its least value avoids square roots.

Step-by-Step Solution

1. $\Delta x = (t+1) - t = 1$.
2. $\Delta y = (t^3-6t-6) - (t^3-16t-3) = 10t - 3$.
3. $D^2 = 1 + (10t-3)^2$.
4. $(10t-3)^2 \ge 0$, minimised (equal to $0$) at $t = \tfrac{3}{10}$.
5. Then $D^2 = 1 + 0 = 1$, so $D = 1$.

Common Mistakes

• Miscomputing $\Delta y$ by dropping a sign in the cubic terms. …

Concept and Intuition

A point on $y=x+3$ can always be written as $(t,t+3)$. Imposing "distance $2$ from $(0,3)$" turns this into an equation in $t$, which has two solutions (two points on the line satisfy the condition, symmetric about the foot of the perpendicular from $(0,3)$). We then need the one nearer the origin.

Step-by-Step Solution

1. Let the point be $(t, t+3)$.
2. Distance from $(0,3)$: $\sqrt{(t-0)^2 + (t+3-3)^2} = \sqrt{t^2+t^2} = |t|\sqrt2 = 2 \Rightarrow t^2 = 2$.
3. Distance-squared from the origin: $t^2 + (t+3)^2 = 2t^2+6t+9$.
4. Using $t^2=2$: this is $4+6t+9 = 13+6t$.
5. For $t=\sqrt2$: $13+6\sqrt2$. For $t=-\sqrt2$: $13-6\sqrt2$. …

Concept and Intuition

For a point outside a circle, the shortest distance to the circle (along the line joining the point to the center) is the distance to the center minus the radius.

Step-by-Step Solution

1. $x^2+y^2-4x-2y-20=0\Rightarrow$ center $(2,1)$, radius $=\sqrt{2^2+1^2+20}=\sqrt{25}=5$.
2. Distance from $(10,7)$ to $(2,1)$: $\sqrt{(10-2)^2+(7-1)^2}=\sqrt{64+36}=\sqrt{100}=10$.
3. Since $10>5$, the point is outside the circle. Least distance $=10-5=5$.

Common Mistakes …

Concept and Intuition

The shortest distance from a curve to a line is measured along the perpendicular to the line. At the closest point on the curve, the curve's tangent must be parallel to the line — otherwise you could slide along the curve a little and get closer.

Step-by-Step Solution

1. Curve: $y=x^2+4x+3$, so $y'=2x+4$.
2. The line $y=3x+2$ has slope $3$. Set the tangent slope equal: $2x+4=3 \Rightarrow x=-\dfrac12$.
3. Find $y$: $y=\left(-\frac12\right)^2+4\left(-\frac12\right)+3 = \frac14-2+3 = \frac54$.
4. Point: $\left(-\dfrac12,\dfrac54\right)$. …

Concept and Intuition

Minimising a distance is always easier if you minimise the squared distance instead (avoids a messy square root and doesn't change where the minimum occurs, since squaring is monotonic for non-negative values). For a point on $y=x^2-4$, the squared distance to the origin is a function purely of $x^2$, so substituting $u=x^2\ge0$ turns it into a simple one-variable quadratic minimisation.

Step-by-Step Solution

1. A general point on the curve is $(x,x^2-4)$. Squared distance from origin: $D^2=x^2+(x^2-4)^2$.
2. Let $u=x^2\ge0$: $D^2=u+(u-4)^2=u+u^2-8u+16=u^2-7u+16$.
3. This is an upward parabola in $u$; minimum at $u=\dfrac{7}{2}=3.5$, which is $\ge0$, so it's attainable. …

Concept and Intuition

Among all lines through a fixed point $P$, the distance from another fixed point $Q$ to the line is at most $|PQ|$ (achieved only when the line is perpendicular to $PQ$ at $P$ — any other orientation makes the foot of perpendicular from $Q$ fall short of the full segment length). So we don't need calculus — just find the perpendicular direction.

Step-by-Step Solution

1. $P=(1,2)$, $Q=(3,1)$. Slope of $PQ$: $\dfrac{1-2}{3-1}=\dfrac{-1}{2}=-\dfrac12$. …