Question

If the tangents drawn to the parabola at the extremities of a common chord $AB$ of the circle $x^2+y^2=5$ and the parabola $y=b{x}^2$ intersect at the point $T$ which lies on the directrix of the parabola, then $\frac{1}{b}=$

• A. $2$
• B. $-2$
• C. $4$
• D. $-4$

Answer 1

Answer: (C), (D)

The correct options are
C $4$
D $-4$


$x^2=\frac{y}{b}$
Comparing with $x^2=4ay$
$\Rightarrow a=\frac{1}{4b}$ and Directrix $:y=\frac{-1}{4b}$
Parametric coordinates of $A$ is
$(2at,a{t}^2)=(\frac{t}{2b},\frac{t^2}{4b})$

Since, $A$ and $B$ lie on the circle also,
coordinates of $B$ is $(\frac{-t}{2b},\frac{t^2}{4b})$

$y=b{x}^2\Rightarrow y^{\prime }=2bx$
Slope at point $A$, $y_A^{\prime }=t$
Slope at point $B$, $y_B^{\prime }=-t$

Equation of tangents at $A$ and $B$ are
$(y-\frac{t^2}{4b})=t(x-\frac{t}{2b})\cdots \left(1\right)$
$(y-\frac{t^2}{4b})=-t(x+\frac{t}{2b})\cdots \left(2\right)$

From $\left(1\right)$ and $2$, we get
$x=0,y=\frac{-t^2}{4b}$
$\therefore \frac{-t^2}{4b}=\frac{-1}{4b}\Rightarrow t=\pm 1$

Therefore, points $A$ and $B$ are $(\pm \frac{1}{2b},\frac{1}{4b})$
$A$ and $B$ lie on the circle $x^2+y^2=5$
$\Rightarrow \frac{1}{4b^2}+\frac{1}{16b^2}=5$
$\Rightarrow b=\pm \frac{1}{4}$

$\Rightarrow b=\frac{1}{4}$ (Upward parabola)
and $b=-\frac{1}{4}$ (Downward parabola)

$\therefore \frac{1}{b}=\pm 4$