Question

The parabolas $C_{1}:y^{2}=4a\left(x - a\right)$ and $C_{2}:y^{2}=-4a\left(x - k\right)$ intersect at two distinct points $A$ and $B.$ If the slope of the tangent at $A$ on $C_{1}$ is same as the slope of the normal at $B$ on $C_{2}$ , then the value of k is equal to

• A. $3a$
• B. $2a$
• C. $a$
• D. $0$

Answer 1

Answer: (A)

Let, $A=(p, q) \& B=(p,-q)$
Now $q^{2}=4 a(p-a) \& q^{2}=-4 a(p-k) \ldots$ (i)
$\Rightarrow p-a=-p+k \Rightarrow p=\frac{a+k}{2} \ldots$
Now for $C_{1}, \frac{d y}{d x}=\frac{2 a}{y}$
$\left.\left.\frac{d y}{d x}\right]_{(p, q)}=\frac{2 a}{q} \&-\frac{d x}{d y}\right]_{(p,-q)}=\frac{q}{2 a}$
$\Rightarrow \frac{2 a}{q}=\frac{q}{2 a} \Rightarrow q^{2}=4 a^{2} \ldots$ (iii)
Putting values from (ii) \& (iii) in equation (i), we get, $4 a^{2}=4 a\left(\frac{a+k}{2}-a\right) \Rightarrow a=\frac{k-a}{2} \Rightarrow k=3 a$