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Q.
If the tangent at the point $P \left( x _1, y _1\right)$ to the parabola $y ^2=4 ax$ meets the parabola $y^2=4 a(x+b)$ at $Q$ \& $R$, then the mid point of $Q R$ is :
Conic Sections
Solution:
chord of the parabola
$y^2=4 a(x+b) \text { whose middle point is }(h, k) $
$y-k=m(x-h) $.....(1)
$\text { where } m=\frac{y^{\prime \prime}-y^{\prime}}{x^{\prime \prime}-x^{\prime}}$
Now $\frac{y^{2 \prime \prime}-y^{2 \prime}}{x^{\prime \prime}-x^{\prime}}=4 a$; but $x^{\prime}+x^{\prime \prime}=2 h$ and $y^{\prime}+y^{\prime \prime}=2 k$
$\frac{y^{\prime \prime}-y^{\prime}}{x^{\prime \prime}-x^{\prime}}=\frac{4 a}{y^{\prime \prime}+y^{\prime}}=\frac{4 a}{2 k}=\frac{2 a}{k}$
Hence (1) becomes
$y-k=\frac{2 a}{k}(x-h)$
$k y-k^2=2 a x-2 a h $
$2 a x-k y+k^2-2 a h=0$.....(2)
Equation of tangent at $P \left( x _1, y _1\right)$ on $y ^2=4 ax$
$yy _1=2 a \left( x + x _1\right) $
$2 ax - yy _1+2 ax _1=0$.....(3)
comparing (2) and (3)
$1=\frac{ k }{ y _1}=\frac{ k ^2-2 ah }{2 ax _1}$
$\Rightarrow k = y _1 \text { and } 2 ax _1= k ^2-2 ah = y _1^2-2 ah =4 ax _1-2 ah \left( as k = y _1\right) $
$\therefore ah =2 ax _1 \Rightarrow h = x _1 $
$\text { Hence } h = x _1 \text { and } k = y _1$
