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Q.
Three normals are drawn from the point $(14,7)$ to the parabola $y^{2}-16 x-8 y=0$. The coordinates of the feet of the normals are
Conic Sections
Solution:
Given parabola is $y^{2}-16 x-8 y=0 ...$(1)
Let the coordinates of the feet of the normal from $(14,7)$ be $P(\alpha, \beta)$
Now, equation of the tangent at $P(\alpha, \beta)$ to parabola (1) is $y \beta$ $-8(x+\alpha)-4(y+\beta)=0$
or, $(\beta-4) y=8 x+8 \alpha+4 \beta ....$(2)
Its slope $=\frac{8}{\beta-4}$.
Equation of normal to parabola (1) at $(a, b)$ is
$y-\beta=\frac{4-\beta}{8}(x-\alpha)$
It passes through $(14,7)$,
$\therefore 7-\beta=\frac{4-\beta}{8}(14-\alpha)$ or $\alpha=\frac{6 \beta}{\beta-4} ...$(3)
Also, $(\alpha, \beta)$ lies on parabola (1),
$\therefore \beta^{2}-16 \alpha-8 \beta=0\,\,\, $(4)
Putting the value of $a$ from (3) in (4), we get
$\beta^{2}-\frac{96 \beta}{\beta-4}-8 \beta=0$
$ \Rightarrow \beta\left(\beta^{2}-12 \beta-64\right)=0$
$\Rightarrow \beta(\beta-16)(\beta+4)=0 $
$\therefore \beta=0,16,-4 .$
From (3), when $\beta=0, \alpha=0$,
when $\beta=16, \alpha=8$ and when $\beta=-4, \alpha=3$.
Hence, the feet of the normals are $(0,0),(8,16)$ and $(3,-4)$.