Question

Let $A$ and $B$be two distinct points on the parabola $y^2=4x.$ If the axis of the parabola touches a circle of radius $2$ having $AB$ as its diameter, then the slope of the line joining $A$ and $B$ can be

• A. $–\left(\frac{1}{2}\right)$
• B. $\frac{1}{2}$
• C. $1$
• D. None of these

Answer 1

The correct option is C

$1$

Explanation for the correct option:

Finding the slope of the line joining $A$ and $B$:

Given: $A$ and $B$are two distinct points on the parabola $y^2=4x.$ Comparing this with the equation $y^2=4ax$, we get $a=1$.

The axis of the parabola touches a circle of radius $2$ having $AB$ as its diameter.

Let us consider the two points to be $A(a{t_1}^2,2a{t}_1)\text{and}B(a{t_2}^2,2a{t}_2)$

Now as $a=1$,

The midpoint formula is given by $\mathbf{}\left[\frac{(x_1+x_2)}{2},\frac{(y_1+y_2)}{2}\right]$

Therefore the mid-point of

$AB=\left[\frac{({t_1}^2+{t_2}^2)}{2},\left(t_1+t_2\right)\right]$

The equation of the circle is ${\left[x–h\right]}^2+{[y–k]}^2=r^2$

$\Rightarrow {\left[x–\frac{({t_1}^2+{t_2}^2)}{2}\right]}^2+{[y–(t_1+t_2\left)\right]}^2=2^2$

$\Rightarrow x^2+y^2–2x\left[\frac{({t_1}^2+{t_2}^2)}{2}\right]–2y(t_1+t_2)+{\left[\frac{({t_1}^2+{t_2}^2)}{2}\right]}^2+{(t_1+t_2)}^2–4=0$

Since it touches the x-axis, $g^2=c$

$\Rightarrow g=\frac{({t_1}^2+{t_2}^2)}{2},c={\left[\frac{({t_1}^2+{t_2}^2)}{2}\right]}^2+{\left[\right(t_1+t_2\left)\right]}^2–4$

$\Rightarrow (t_1+t_2)=2....\left(1\right)$

Hence the Slope

$AB=\frac{[2(t_1+t_2\left)\right]}{({t_2}^2–{t_1}^2)}$

$$\begin{gathered} =\frac{2}{(t_1+t_2)} \\ =\frac{2}{2} \\ =1 \end{gathered}$$ From equation $\left(1\right)$

Hence, option (C) is the correct answer.