Question

Let P : y 2 = 4 ax , a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of π/4 with the line y = 3 x + 5 touch the parabola P at A and B. Then the value of a for which A , B and S are collinear is

• A. 8 only
• B. 2 only
• C. $\frac{1}{4}$ only
• D. any a > 0

Answer 1

Answer: (D)

any a > 0

Answer 2

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