\(\frac{1}{4}\) only
The correct answer is (D) : any a > 0
\(P:y²=4ax, a > 0 \) \(S(a,0)\)
Equation of tangent on parabola
\(y = mx + \frac{a}{m}\)
y = 3x + 5
\(tan \frac{π}{4} = |\frac{m-3}{1+3m} | ⇒ m-3 = ± ( 1+3m )\)
m-3 = 1+3m
m=-2
m-3 = -1-3m
\(m=\frac{1}{2}\)
Equation of one tangent :\( y = -2x - \frac{a}{2}\)
Equation of other tangent : \(y = \frac{x}{2} + 2a\)
Point of contact are
\(( \frac{a}{(-2)²} , \frac{-2a}{(-2)} ) and ( \frac{a}{(\frac{1}{2})²} , \frac{-2a}{\frac{1}{2}} )\)
\(A ( \frac{a}{4},a)\) and \(B (4a,-4a)\)
Now or \((ΔABS)\) = 0 [ S is the focus ]
\(\frac{1}{2} \begin{vmatrix} \frac{a}{4} & a & 1 \\ 4a & -4a & 1 \\ a& 0 & 1 \end{vmatrix} = 0\)
\(⇒ \frac{a}{4} (-4a-0) -a(4a-a) + 1(0-(-4a²)) = 0\)
\(= -a² -3a² + 4a² = 0\)
Always true
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:
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 one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) 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