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:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):

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