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 $q^2=4a(p-a)\&q^2=-4a(p-k)\dots$ (i)
$\Rightarrow p-a=-p+k\Rightarrow p=\frac{a+k}{2}\dots$
Now for $C_1,\frac{dy}{dx}=\frac{2a}{y}$
${{\frac{dy}{dx}]}_{(p,q)}=\frac{2a}{q}\&-\frac{dx}{dy}]}_{(p,-q)}=\frac{q}{2a}$
$\Rightarrow \frac{2a}{q}=\frac{q}{2a}\Rightarrow q^2=4a^2\dots$ (iii)
Putting values from (ii) \& (iii) in equation (i), we get, $4a^2=4a\left(\frac{a+k}{2}-a\right)\Rightarrow a=\frac{k-a}{2}\Rightarrow k=3a$