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
\(=\frac{\frac{1}{4}+6}{\frac{1}{2}+2}=\frac{5}{2}\)
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
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:
Let the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ 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