Question

Two mutually perpendicular tangents of the parabola $y^2=4ax$ meet its axis in $P_1$ and $P_2$. If S is the focus of the parabola then $\frac{1}{SP_1}+\frac{1}{SP_2}$ is equal to :

• A. $\frac{4}{a}$
• B. $\frac{2}{a}$
• C. $\frac{1}{a}$
• D. $\frac{1}{4a}$

Answer 1

Answer: (C)

$\frac{1}{a}$

Answer 2

The focus of the parabola $y^2=4ax$ is $S=(a,0)$. The axis is the x-axis. The equation of a tangent to the parabola in parametric form is $yt = x + at^2$. When this tangent meets the axis ($y=0$), we have $0 = x + at^2$, so $x = -at^2$. Let the points where the two perpendicular tangents meet the axis be $P_1 = (-at_1^2, 0)$ and $P_2 = (-at_2^2, 0)$. The slope of the tangent $x - yt + at^2 = 0$ is $m = \frac{1}{t}$. For perpendicular tangents, $m_1 m_2 = -1$, which means $\frac{1}{t_1} \cdot \frac{1}{t_2} = -1$, so $t_1 t_2 = -1$. This implies $t_2 = -\frac{1}{t_1}$, and thus $t_2^2 = \frac{1}{t_1^2}$. The distances from the focus $S(a,0)$ to $P_1$ and $P_2$ are: $SP_1 = |a - (-at_1^2)| = |a(1+t_1^2)|$ $SP_2 = |a - (-at_2^2)| = |a(1+t_2^2)| = |a(1+\frac{1}{t_1^2})| = |a\frac{t_1^2+1}{t_1^2}|$ We need to find $\frac{1}{SP_1} + \frac{1}{SP_2}$: $\frac{1}{SP_1} = \frac{1}{|a(1+t_1^2)|}$ $\frac{1}{SP_2} = \frac{1}{|a\frac{t_1^2+1}{t_1^2}|} = \frac{t_1^2}{|a(t_1^2+1)|}$ $\frac{1}{SP_1} + \frac{1}{SP_2} = \frac{1}{|a(1+t_1^2)|} + \frac{t_1^2}{|a(t_1^2+1)|} = \frac{1+t_1^2}{|a(1+t_1^2)|}$ Since $1+t_1^2 > 0$, this simplifies to $\frac{1+t_1^2}{|a|(1+t_1^2)} = \frac{1}{|a|}$. Assuming $a>0$ (as is standard and implied by the options), the result is $\frac{1}{a}$.