Question

Three normals are drawn from the point $(14,7)$ to the parabola $y^2-16x-8y=0$. The coordinates of the feet of the normals are

• A. $(0,0),(8,-16),(3,-4)$
• B. $(0,0),(8,16),(3,-4)$
• C. $(0,0),(-8,16),(3,-4)$
• D. none of these

Answer 1

Answer: (B)

Given parabola is $y^2-16x-8y=\mathrm{0...}$(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, $(\beta -4)y=8x+8\alpha +4\beta ....$(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),
$\therefore {\beta }^2-16\alpha -8\beta =0$
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 $(3,-4)$.