Animated Solution for Mathematics - Conic Sections: If the line $y=4+kx,k>0$, is the tangent to the parabola $y=x-x^2$ at the point $P$ and $V$ is the vertex of the parabola, then the slope of the line through $P$ and $V$ is :

A Journey Through Tangency

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a beautiful problem in coordinate geometry. It is not just about finding a slope; it is about understanding the hidden geometry of a parabola and the rigid constraints that define its tangents.

We begin with the parabola $y=x-x^2$. The negative coefficient of $x^2$ tells us immediately that our parabola is a "sad" one—it opens downwards.

If we set $y=0$, we find the roots at $x=0$ and $x=1$. To find the peak, we identify the vertex $V$ using the formula $x_v=-\frac{b}{2a}$.

With $a=-1$ and $b=1$, we find:

Plugging this back into our equation, we get $y_v=\frac{1}{2}-(\frac{1}{2}{)}^2=\frac{1}{4}$. Thus, our vertex $V$ is anchored at $(\frac{1}{2},\frac{1}{4})$.

Now, consider the tangent line $y=kx+4$. We are told this line touches the parabola at some point $P(\alpha ,\beta )$. Since $P$ sits on the parabola, its coordinates must satisfy the equation: $\beta =\alpha -{\alpha }^2$.

To find the slope of the tangent at $P$, we use the derivative $\frac{dy}{dx}=1-2x$. At our point $P$, where $x=\alpha$, the slope is $k=1-2\alpha$.

Using the point-slope form $y-\beta =k(x-\alpha )$, we substitute $\beta =\alpha -{\alpha }^2$ and $k=1-2\alpha$:

Expanding this, we get $y=(1-2\alpha )x-\alpha +2{\alpha }^2+\alpha -{\alpha }^2$, which simplifies to:

We now have two expressions for the same line: the one given in the problem, $y=kx+4$, and our derived equation, $y=(1-2\alpha )x+{\alpha }^2$. By comparing the constant terms, we see that ${\alpha }^2=4$, which gives us two candidates for $\alpha$: $2$ and $-2$.

We must respect the constraint $k>0$. If $\alpha =2$, then $k=1-2(2)=-3$, which is negative and must be rejected.

If $\alpha =-2$, then $k=1-2(-2)=5$, which is positive. This is our winner! With $\alpha =-2$, our point of tangency $P$ is $(-2,-6)$.

We have the vertex $V(\frac{1}{2},\frac{1}{4})$ and the point of tangency $P(-2,-6)$. The slope $m$ of the line passing through these two points is calculated as:

Simplifying this expression, we get:

The final slope is $\frac{5}{2}$. Remember, the beauty of these problems lies in the logical steps that lead you there; always check your constraints and trust your algebra.