The correct option is(C): \(\frac{5}{2}\).
∵ Line y = kx + 4 touches the parabola y = x – x2.
So, kx + 4 = x – x2 ⇒ x2 + (k – 1) x + 4 = 0 has only one root
(k – 1)2 = 16 ⇒ k = 5 or – 3 but k > 0
So, k = 5.
And hence x2 + 4x + 4 = 0 ⇒ x = – 2
So, P(–2, –6) and V is
\((\frac{1}{2},\frac{1}{4})\)
Slope of PV
\(=\frac{\frac{1}{4}+6}{\frac{1}{2}+2}=\frac{5}{2}\)
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
=> MP2 = PS2
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2